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PHP - Getting Image From Database
Ok i have been trying for a few days now with no success to show a image from my database but as this is the first time i have tried to use databases im having no luck, my problem is this...
I have an image gallery that shows a list of differnt galleries with the number of images inside it, like this.... General (10) I have a form that allows me to add new galleries to the list and upload an image to be shown in the gallery list, i simply added a field in the categories table and called it category_img. I would like to make the list of galleries look better by having an image for each one like this.... This is the code that generates and displays the list of galleries/categories but it is WAY to advanced and confusing for me to work out what to do, i have tried all sorts but only ever destroys what i already have or just does nothing at all, the closest i have got was displaying the name of the image, but no actual image. Code: [Select] <?php // initialization $result_array = array(); $counter = 0; $cid = (int)($_GET['cid']); $pid = (int)($_GET['pid']); // Category Listing if( empty($cid) && empty($pid) ) { $number_of_categories_in_row = 1; $result = mysql_query( "SELECT c.category_id,c.category_name,COUNT(photo_id) FROM gallery_category as c LEFT JOIN gallery_photos as p ON p.photo_category = c.category_id GROUP BY c.category_id" ); while( $row = mysql_fetch_array( $result ) ) { $result_array[] = '<p><a href="gallery.php?cid=' . $row[0] . '">' . $row[1] . '</a> (' . $row[2] . ')</p>'; } mysql_free_result( $result ); $result_final = "<tr>\n"; foreach($result_array as $category_link) { if($counter == $number_of_categories_in_row) { $counter = 1; $result_final .= "\n</tr>\n<tr>\n"; } else $counter++; $result_final .= "\t<td>".$category_link."</td>\n"; } if($counter) { if($number_of_categories_in_row-$counter) $result_final .= "\t<td colspan='".($number_of_categories_in_row-$counter)."'> </td>\n"; $result_final .= "</tr>"; } } i could really do with some help as trying to learn this stuff is really hard when i have to work in my living room with children running around, my head has not stopped hurting in days. Similar TutorialsHi, I've read a lot of places that it's not recommended to store binary files in my db. So instead I'm supposed to upload the image to a directory, and store the link to that directory in database. First, how would I make a form that uploads the picture to the directory (And what kinda directories are we talking?). Secondly, how would I retrieve that link? And I guess I should rename the picture.. I'd appreciate any help, or a good tutorial (Haven't found any myself). Hello, Five images will be displayed inside a division. There will be a previous and next button/link. If someone click the next button the next image will be added in that div and the first image will be gone from that div. The previous button/link will do the same thing. Is it possible with php? I am confused if it's a javascript or ajax question. Thanks. hello friends, while clicking the form all the information goes to database, I have one image upload field, when cliking the submit button, i would like 'image name' to go in database and file to go in /upload folder, i have tried this for hours and gave up, if anyone help me in this, i would be very greatful Hi, I have a search results page that return the results based on "Make" & "Model" of vehicle. The question is, can an image be echoed onto that results page based on which make and model has been searched for? The image is stored in a field called "Image_Van" so would need to display that depending on the row in that database that has been selected from Make & Model... Hope this makes sense! <?php //info bars colour $color_info="#D6D685"; // Almost Silver //FFFFCC $color1="#E9E9E9";// Light Blue $color2="#EAD5FF";// Light Blue //include function library code include_once("func_lib.php"); // use the user_connection include file's connectDB function include_once("usr_conn.php"); if (!connectDB()) { echo "<p>Unable To Connect To Database</p>"; return; } //test that make and model are passed in, if they are assign them if (($_GET ['make'] == "MAKE") or ($_GET ['model'] == "MODEL")) { $errMessage = "<br><br><b><font class=small>Your search hasn't returned any results, please re-enter your search again.</font></b><br><br><a href=van_accessories.php>Search again</a><br><br>"; } else { // get search values for make and model, removing " and ' chars $strMake = mysql_escape_string($_GET ['make']); $strModel = mysql_escape_string($_GET ['model']); } // set the Prod_Type that is to be shown regardless of car make and model $strDefaultProd = "XD"; //write query string that will pased in paging links $strQuery = "make=".$strMake."&model=".$strModel."&"; //echo $strQuery; //grab the page number that we're on $page = mysql_escape_string($_GET ['page']); //$strMake = "FIAT"; //$strModel = "Uno"; //create sql query $sqlSelect = "SELECT Prod_ID, Prod_Type, Prod_Model, Prod_Make, Prod_REF, Product_Desc, Price_ExVat, image_name, Image_Van FROM products"; //change this back to 'FILL THIS IS IN' to switch accessories off! $criteria = " WHERE (Car_Make = '".$strMake."' AND Car_Model = '".$strModel."' AND Default_Code = 'RS') OR (Car_Make = 'all' AND Default_Code = 'SA')"; //$criteria = " WHERE Car_Make = '".$strMake."' AND Car_Model = '".$strModel."'"; // assign the basic sqlquery $sqlquery = $sqlSelect . $criteria; //echo $sqlquery; #debug //get the result set $result = mysql_query($sqlquery); $count = mysql_num_rows($result); $full_count = $count; //set max number of records per page $records_per_page = 999; //Next figure out how many pages you'll have. $total_pages = ceil( $count / $records_per_page ); /* This just sets up a page variable which will be past in a "next" and "prev" link. The first link to this page may not actually have $page so this just sets it up. */ if ( $page == "" ) { $page = 1; } //echo "<br>page number".$page; #debug # this will set up the "First | Prev | " part of our navigation. if ( $page == 1 ) { # if we are on the first page then "First" and "Prev" # should not be links. $naviagation = "<font color=\"#666666\">First</font> | <font color=\"#666666\">Prev</font> | "; } else { # we are not on page one so "First" and "Prev" can # be links $prev_page = $page - 1; $navigation = "<a href=\"van_steps_results.php?".$strQuery."page=1\">First</a> | <a href=\"van_steps_results.php?".$strQuery."page=".$prev_page."\">Prev</a> | "; } # this part will set up the rest of our navigation "Next | Last" if ( $page == $total_pages ) { # we are on the last page so "Next" and "Last" # should not be links // $navigation .= "<font color=\"#666666\">Next</font> | <font color=\"#666666\">Last</font>"; } else { # we are not on the last page so "Next" and "Last" # can be links $next_page = $page + 1; $navigation .= "<a href=\"van_steps_results.php?".$strQuery."page=".$next_page."\">Next</a> | <a href=\"van_steps_results.php?".$strQuery."page=".$total_pages."\">Last</a>"; } /* The final thing to do is your actual query, but first we have to figure out what the offset will be. */ $offset = ( $page - 1 ) * $records_per_page; //echo "<br>offset".$offset; #debug //create sql query $sqlSelect = "SELECT Car_Make, Car_Model, Prod_ID, Prod_Type, Prod_Model, Prod_Make, Priority, Prod_Code, Prod_REF, Product_Desc, Price_ExVat, Link, image_name, Image_Van FROM products"; $criteria = " WHERE (Car_Make = '".$strMake."' AND Car_Model = '".$strModel."' AND Default_Code = 'RS') OR (Car_Make = 'all' AND Car_Model = 'all' AND Default_Code = 'SA')"; $limit = " ORDER BY Priority ASC, Prod_Code ASC, Prod_ID ASC LIMIT $offset, $records_per_page"; // assign the basic sqlquery $sqlquery = $sqlSelect . $criteria . $limit ; //echo $sqlquery; //get the result set $result = mysql_query($sqlquery); // if no matches found from query if (!$result || (mysql_num_rows($result) == 0)){ //echo ("<b><font class=small>No matches were found.</font></b>"); } // display the results else{ // use rowcount to display total results found $count = mysql_num_rows($result); //echo $count; //echo("<p>Total Matches Found = " . $rowCount . "</p>"); //*debug } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html> <head> <title>Results Page:</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <script src="js/jquery-1.3.2.min.js" type="text/javascript"></script> <!--script src="js/jquery.lint.js" type="text/javascript" charset="utf-8"></script--> <link rel="stylesheet" href="css/prettyPhoto.css" type="text/css" media="screen" title="prettyPhoto main stylesheet" charset="utf-8" /> <script src="js/jquery.prettyPhoto.js" type="text/javascript" charset="utf-8"></script> <!-- start header graphic html --> <style type="text/css"> <!-- .table { border: 0px none; border-collapse: separate; } BODY { } A { COLOR: #003366; TEXT-DECORATION: none; } A:hover { TEXT-DECORATION: underline; color: #900; } .small { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; } .std { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; COLOR: #000000; } .header { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; FONT-WEIGHT: BOLD; COLOR:#000000; } .bigheader { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; FONT-WEIGHT: BOLD; COLOR:#FFFFFF; } .error { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; COLOR:#FF0000 } .headertable { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; FONT-WEIGHT: regular; COLOR:#FFFFFF; BACKGROUND-COLOR:#000066; } .stdtable { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; FONT-WEIGHT: regular; COLOR:#000066; } .style16 {font-size: 11px} .style7 { color: #CC0000; font-style: italic; } .small1 { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; } .header1 { FONT-SIZE: 8pt; FONT-FAMILY: Verdana; FONT-WEIGHT: BOLD; COLOR:#000000; } a:link { color: #900; } --> </style> <div align="center"> <table border="0" cellpadding="0" cellspacing="0"> <tr> <td background=""><table width="100%" border="0" align="center" cellpadding="2" cellspacing="0"> <tr> <td bgcolor="#003366"> </td> </tr> </table> <!--END header--> <table width="960" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td><table width="100%" border="0" cellspacing="0" cellpadding="2"> <tr> <td><div align="center"> <table width="100%" border="0" cellspacing="0" cellpadding="0"> <tr> <td class="arial11"><table width="100%" border="0" cellpadding="10" cellspacing="0"> <tr> <td width="33%"><p><img src="2003/results.gif" alt="" width="238" height="43" /></p> <p> </p> <p> </p></td> </tr> </table></td> </tr> </table> </div></td> </tr> </table> <table width="950" border="0" align="center" cellpadding="2" cellspacing="2" bordercolor="#FFFFFF"> <tr align="center" valign="top" class="std"> <td colspan="7"><hr width="500" size="1" /> <?php if (isset($errMessage)){ echo $errMessage; } // if no matches found from query if (!$result || (mysql_num_rows($result) == 0)) echo ("<b><font class=small>No matches were found.</font></b><br><br>"); // display the results else{ // use rowcount to display total results found echo("<br><p><em><font class=vehicle>Search Results for " . $strMake . " " . $strModel .".</em> Total Matches Found = " . $full_count . "</p>"); //*debug echo $navigation; } /* With your navigation built you can just echo that out any where you want it to appear. */ //echo $navigation; ?></td> </tr> <?php $title = "<tr align=\"center\" class=\"headertable\" height=\"32\">"; $title .= "<td width=\"95\"><img src=\"images/transparent.gif\" width=\"95\" height=\"1\" /></td>"; $title .= "<td width=\"60\">Make</td>"; $title .= "<td width=\"60\">Ref No.</td>"; $title .= "<td width=\"550\">Title/Description</td>"; $title .= "<td width=\"73\">Price Ex VAT</td>"; $title .= "<td width=\"73\">Price Inc VAT</td>"; $title .= "<td width=\"60\"><img src=\"2003/basket.gif\" width=\"21\" height=\"14\"></td>"; $title .= "</tr>"; //set the counters for the records results display screen $cnt = 0; $headercount1 = 0; $headercount2 = 0; $headercount3 = 0; $headercount4 = 0; $headercount5 = 0; $headercount6 = 0; $headercount7 = 0; while ($row = mysql_fetch_assoc($result)) { //cut the prod_type variable //get the length of the variable $strLength = strlen ($row["Prod_Type"]); //assign first 2 characters of variable $strPrefix = substr($row["Prod_Type"], 0, 2); //debug //echo "\n prefix :" . $strPrefix; //assign remaining characters of variable $strSuffix = substr($row["Prod_Type"], 2, $strLength); //debug //echo "\n suffix :" . $strSuffix; //set product header image depending on the Prod_Type Code //start the table row echo "<tr align=center><td colspan=7 class=small>"; //set bg cell color diff for BH make switch ($strPrefix) { case "BH": $color=$color2; break; default: $color=$color1; break; }//end switch switch ($strPrefix) { case "BH": if ($headercount1 == 0) { echo "<img border=0 src=2003/searchheaders/step.jpg></td></tr>"; echo $title; } $headercount1 ++; //$headercount1 break; // REMOVE ALL THIS COMMENTING IF NEEDED TO SHOW THE ROOF BARS! case "RB": // if ($headercount2 == 0) // { // echo "<img border=0 src=2003/searchheaders/NOTUSED.jpg alt=Roof-Bars></td></tr>"; // echo $title; // } // $headercount2 ++; //$headercount1 break; case "BX": if ($headercount3 == 0) { echo "<img border=0 src=2003/searchheaders/bulkheads.jpg alt=Roof-Boxes></td></tr>"; echo $title; } $headercount3 ++; //$headercount1 break; case "BR": if ($headercount4 == 0) { echo "<img border=0 src=2003/searchheaders/bulkheads.jpg alt=Bike-Racks></td></tr>"; echo $title; } $headercount4 ++; //$headercount1 break; case "SR": if ($headercount5 == 0) { echo "<img border=0 src=2003/searchheaders/bulkheads.jpg alt=Ski-Rack></td></tr>"; echo $title; } $headercount5 ++; //$headercount1 break; case "CR": if ($headercount6 == 0) { echo "<img border=0 src=2003/searchheaders/bulkheads.jpg alt=Canoe-Rack></td></tr>"; echo $title; } $headercount6 ++; //$headercount1 break; case "XD": if ($headercount7 == 0) { echo "<img border=0 src=2003/searchheaders/towbar.jpg alt=General-Accessories></td></tr>"; echo $title; } $headercount7 ++; } //end header image selection and close the table row. //echo "</td></tr>"; // if (($cnt++)%2) // { // $color="#E9E9E9"; // Almost Silver // } // else // { // $color="#E9E9E9";// Light Blue // } //print a product row if the prodtype isn't set to INFO $pop_link = ""; if ($strSuffix != "INFO"){ //test to see if an image exists for the product if ($row["Link"] <> NULL){ $pop_link = " <a href=\"products/".$row["Link"]."?iframe=true&width=650&height=500\" rel=\"prettyPhoto\" >More information >>>></a>"; } echo "<tr class=stdtable align=center height=25>"; //DISPLAY THUMB IMAGE IF ONE WAS LISTED IN THE image_name FIELD $p_Image_Van = htmlspecialchars($row["Image_Van"]); //echo "p_Image_Van:$p_image_name"; if ($p_Image_Van != "") echo "<td bgcolor=$color width=\"95\"><img src=\"images/product_images/$p_Image_Van\" border=\"0\" /></td>"; else echo "<td bgcolor=$color width=\"95\"><img src=\"images/transparent.gif\" width=\"95\" height=\"50\" border=\"0\" /></td>"; echo "<td bgcolor=$color width=\"60\">". htmlspecialchars($row["Prod_Make"]) . "</td><td bgcolor=$color width=\"60\">" . htmlspecialchars($row["Prod_REF"]) . "</td><td bgcolor=$color align=left class=\"infolink\" width=\"550\">" . htmlspecialchars($row["Product_Desc"]) . $pop_link . "</td><td bgcolor=$color width=\"73\">£" . number_format(htmlspecialchars($row['Price_ExVat']), 2) . "</td><td bgcolor=$color width=\"73\">£" . number_format((calcVAT (htmlspecialchars($row['Price_ExVat']))), 2) . "</td><td bgcolor=$color width=\"60\"><a href=basket.php?src=".urlencode($_SERVER['REQUEST_URI'])."&productID=" . $row["Prod_ID"] . "><img src=2003/buy.gif border=0></a></td></tr>"; } // if the prod type is set to INFO then print out a new row for that information else { echo "<tr class=stdtable align=center height=25 ><td colspan=7 bgcolor=$color_info><em><strong>" . htmlspecialchars($row["Product_Desc"]) . "</strong></em></td></tr>"; } } //****************************************************************************************************** /* //get all porducts from the database that are marked with prod_type XD regardless of car make and model $sqlSelectDefault = "SELECT Prod_ID, Prod_Type, Prod_Model, Prod_Make, Priority, Prod_Code, Prod_REF, Product_Desc, Price_ExVat FROM products"; $criteriaDefault = " WHERE Prod_Type LIKE '".$strDefaultProd."%'"; $limitDefault = "ORDER BY Priority ASC, Prod_Code ASC"; // assign the basic sqlquery $sqlqueryDefault = $sqlSelectDefault . $criteriaDefault . $limitDefault ; //echo $sqlquery; //get the result set $resultDefault = mysql_query($sqlqueryDefault); // if no matches found from query if (!$resultDefault || (mysql_num_rows($resultDefault) == 0)){ //echo ("<b><font class=small>No matches were found.</font></b>"); } // display the results else{ // use rowcount to display total results found $countDefault = mysql_num_rows($resultDefault); //echo $count; //echo("<p>Total Matches Found = " . $rowCount . "</p>"); //*debug } //set product header image depending on the Prod_Type Code //start the table row echo "<tr align=left><td colspan=6 class=small>"; echo "<img border=0 src=2003/searchheaders/general.jpg alt=General-Accessories>"; echo "</td></tr>"; while ($rowDefault = mysql_fetch_assoc($resultDefault)) { //cut the prod_type variable //get the length of the variable $strLengthDefault = strlen ($rowDefault["Prod_Type"]); //assign first 2 characters of variable $strPrefixDefault = substr($rowDefault["Prod_Type"], 0, 2); //debug //echo htmlspecialchars($rowDefault["Prod_ID"]); //echo "\n prefix :" . $strPrefixDefault; //assign remaining characters of variable $strSuffixDefault = substr($rowDefault["Prod_Type"], 2, $strLengthDefault); //debug //echo "\n suffix :" . $strSuffixDefault; if ($strSuffixDefault != "INFO"){ $color="#E9E9E9"; // Almost Silver //print a product row if the prodtype isn't set to INFO echo "<tr class=stdtable align=center height=25 ><td bgcolor=$color>". htmlspecialchars($rowDefault["Prod_Make"]) . "</td><td bgcolor=$color>" . htmlspecialchars($rowDefault["Prod_REF"]) . "</td><td bgcolor=$color align=left>" . htmlspecialchars($rowDefault["Product_Desc"]) . "</td><td bgcolor=$color>?" . number_format(htmlspecialchars($rowDefault['Price_ExVat']), 2) . "</td><td bgcolor=$color>?" . number_format((calcVAT (htmlspecialchars($rowDefault['Price_ExVat']))), 2) . "</td><td bgcolor=$color><a href=basket.php?src=".urlencode($_SERVER['REQUEST_URI'])."&productID=" . $rowDefault["Prod_ID"] . "><img src=2003/buy.gif border=0></a></td></tr>"; } // if the prod type is set to INFO then print out a new row for that information else { echo "<tr class=stdtable align=center height=25 ><td colspan=6 bgcolor=$color><em>" . htmlspecialchars($rowDefault["Product_Desc"]) . "</em></td></tr>"; } } */ ?> <tr align="center"> <td colspan="7" class="small1"></td> </tr> <tr> <!-- <tr align="center" class="headertable" height="32"> <td width="145">Product Name</td> <td width="178">Ref No.</td> <td width="280">Description</td> <td width="73">Price Ex VAT</td> <td width="73">Price Inc VAT</td> <td width="60"><img src="2003/basket.gif" width="21" height="14"> </td> </tr>--> <td colspan="7" height="1" bgcolor="#003366"><img src="colour_spacer.gif" width="756" height="1" /></td> </tr> </table></td> </tr> </table> <p><br /> </p> </p> <div align="center"><br /> </div> <table width="100%" border="0" cellpadding="0" cellspacing="0"> <tr> <td> </td> </tr> </table> <?php //free up some memory mysql_free_result($result); // close the aatabase connection mysql_close(); //include footer code ?> <br /> </table> </html> Ok , so im fairly new to this beeen watching tuts , messing around with coding to see what does what , well i cant seem to get this uploader to send a image to the db it just errors the code is <html> <head> <title>Image Uploader</title> </head> <body> <form method="post" action="index.php" enctype="multipart/form=data"> File: <input type="file" name="image" /> <input type="submit" value="upload" /> </form> <?php mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("test") or die(mysql_error()); //file propities $file = $_FILES['image']['tmp_name']; if (!isset($file)) { echo "Please select an image"; } else { $image = file_get_contents ($_FILES['image']['tmp_name']); $image_name = $_FILES['image']['name']; $image_size = getimagesize($_FILES['image']['tmp_name']); if ($image_size==FALSE) echo "This is not an image file"; else { if (!$insert = mysql_query("INSERT INTO imagedb VALUES ('','$image_name','$image')")) echo "problem uploading image"; else { $lastid = mysql_insert_id(); echo "image uploaded<p>Your image</p><img scr =get.php?id=$lastid>"; } } } ?> </body> </html> thats my index.php then my get.php is <?php mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("test") or die(mysql_error()); $id = addslashes($_REQUEST['id']); $image = mysql_query("SELECT * FROM imagedb WHERE id=$id"); $image = mysql_fetch_assoc($image); $image = $image['image']; header("Content-type: img/jpeg"); echo $image; ?> My database is called 'test' and my table is called 'imagedb' and i have 3 fields 'id'name'image. help would be much appriciated thankyouu anybody has any idea what the code is to put in a main image. So far i was able to find this but it pulls up all the images added in the album. I am only looking for the first image. Thank you. <ul class="thumbs thumbs_nocaptions"> <?php foreach( $this->paginator as $photo ): ?> <li> <a class="thumbs_photo" href="<?php echo $photo->getHref(); ?>"> <span style="background-image: url(<?php echo $photo->getPhotoUrl('thumb.normal'); ?>);"></span> </a> </li> <?php endforeach;?> </ul> Hi, wondering if somebody can tell me where I'm going wrong (I'm new to all of this). I have the following php code which uploads an image file into my database: Code: [Select] //Connect to database include 'Resources/Include/db.inc.php'; $tmp=$_FILES['image']['tmp_name']; //get users IP $ip=$_SERVER['REMOTE_ADDR']; //Don't do anything if file wasn't selected if (!empty($tmp)) { //Copy file to temporary folder copy($tmp, "./temporary/".$ip.""); //open the copied image, ready to encode into text to go into the database $filename1 = "./temporary/".$ip; $fp1 = fopen($filename1, "rb"); //record the image contents into a variable $contents1 = fread($fp1, filesize($filename1)); $contents1 = addslashes($contents1); //close the file fclose($fp1); $ftype = $_FILES['image']['type']; //insert information into the database if(!mysql_query("INSERT INTO LetterImages (Data,Type,LetterID,Page)"." VALUES ( '$contents1', '$ftype',1,1)")){ echo mysql_error(); } //delete the temporary file we made unlink($filename1); } This seems to work ok, as when I go to the LetterImages table there is now an additional row with a file in the blob field. I then have the following code which is supposed to display the image: Code: [Select] $result=mysql_query("SELECT * FROM LetterImages WHERE LetterID=1 AND Page=1"); //fetch data from database $sqldata=mysql_fetch_array($result); $encoded=stripslashes($sqldata['Data']); $ftype=$sqldata['Type']; //tell the browser what type of image to display header("Content-type: $ftype"); //decode and echo the image data echo $encoded; Instead of displaying an image, however, this just displays pages and pages of incomprehensible data. Can anybody tell me where I'm going horribly wrong? the script succesfuly insert image to the database bt, i cant be able to display it on my pages, any help i will appreciate
Attached Files
saveimage.php 1.15KB
2 downloads
images_tbl.php 192bytes
3 downloads hi all i need a script that will allow me upload images to 2 dif folders on my server and then add the info to my database along with some other form data. iv been looking all over for code or scripts for days now and have been playing with cut and copyed code but no look again any help i will be greatful for as im a noob to php but al learning quick here is my html form Code: [Select] <html> <body> <form action="add.php" method="post"> Project Name: <input type="text" name="pro_name" /><br> Thumbnail: <input type="file" name="thumbnail" /><br> ////// this image to ../thum Short Details: <input type="text" name="short_details" /><br> Full Details: <input type="text" name="full_details" /><br> Category: <input type="text" name="cat" /><br> Image1: <input type="file" name="image1" /><br>//// and image1,2,3,4 to ../images Image2: <input type="file" name="image2" /><br> Image3: <input type="file" name="image3" /><br> Image4: <input type="file" name="image4" /><br> <input type="submit" /> </form></body></html> here is my code for add.php witch only adds the info to the DB Code: [Select] <?php error_reporting(E_ALL); include ("../includes/db_config.php"); $con = mysql_connect($db_hostname,$db_username,$db_password); @mysql_select_db($db_database) or die( "Unable to select database"); $sql="INSERT INTO $db_table (pro_name, thumbnail, short_details, full_details, cat, image1, image2, image3, image4) VALUES ('$_POST[pro_name]','$_POST[thumbnail]','$_POST[short_details]','$_POST[full_details]','$_POST[cat]','$_POST[image1]','$_POST[image2]','$_POST[image3]','$_POST[image4]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> Hi there I've been working with some code to display a single record on page. This all works fine and I'm able to pull what I want from the database. My problems is trying to use that data and turning it into something else like a link. I have a field in the database called image url which contains rows of image urls. So here is the problem area of the code: Code: [Select] <?php //////Displaying Data///////////// $id=$_GET['id']; // Collecting data from query string if(!is_numeric($id)){ // Checking data it is a number or not echo "Data Error"; exit; } $fetch=mysql_query("select * from productfeeds where ProductID=$id "); $row=mysql_fetch_object($fetch); echo mysql_error(); echo "<table>"; echo " <tr><td><b>ProductID</b></td><td>$row->ProductID</td></tr> <tr><td><b>ProductName</b></td><td>$row->ProductName</td></tr> <tr><td><b>ProductPrice</b></td><td>$row->ProductPrice</td></tr> //problem area for me <tr><td><b>Image</b></td><td>$row->ImageURL</td></tr> echo "</table>"; I'm trying to edit this part of the code: <tr><td><b>Image</b></td><td>$row->ImageURL</td></tr> I've tried this: <tr><td><b>Image</b></td><td><a href='{$row['URL']}'> <img src='{$row['ImageURL']}'></a> and <tr><td><b>Image</b></td><td><a href='$row['URL']'> <img src='$row['ImageURL']'></a> //removed brackets but I'm just getting errors. Can you guys help please? Thank you very much. hi, i have a form that requires image to be uploaded. i know how to store the form values in a database but i don't know how to store the image into the database. i want the user to upload picture and preview the data provided review (including the image) before submitting to the database. thanks I have a PHP CRUD application and when i delete a row i need it to delete from the MySQL and the image from the uploads folder as well. I have researched and tried dozens of ways with the unlink option but nothing works. If i take out the unlink from my code it will delete fine from the DB. I am new to coding and PHP so any help would be awesome. The file_path is correct. The uploads is the name of the folder where the image is stored and the $_POST["image"] is the column name in MySQL where the image name is stored. delete.php
<?php
//start PHP session
session_start();
if (!isset($_SESSION['success']))
{
header("Location: login_page.php");
die();
}
// check if value was posted
if($_POST){
// include database and object file
include_once 'config/database.php';
$file_path = 'uploads/' . $_POST["image"];
if(unlink($file_path))
{
// delete query
$query = "DELETE FROM myDBname WHERE id = ?";
$stmt = $con->prepare($query);
$stmt->bindParam(1, $_POST['object_id']);
}
if($stmt->execute()){
// redirect to read records page and
// tell the user record was deleted
echo "Record was deleted.";
}else{
echo "Unable to delete record.";
}
}
?>
this is the delete button code
echo "<a delete-id='{$id}' class='btn btn-danger delete-object'>";
echo "<span class='glyphicon glyphicon-remove'></span> Delete";
echo "</a>";
This is the javascript for the delete button as well.
// delete record
$(document).on('click', '.delete-object', function(){
var id = $(this).attr('delete-id');
bootbox.confirm({
message: "<h4>Are you sure?</h4>",
buttons: {
confirm: {
label: '<span class="glyphicon glyphicon-ok"></span> Yes',
className: 'btn-danger'
},
cancel: {
label: '<span class="glyphicon glyphicon-remove"></span> No',
className: 'btn-primary'
}
},
callback: function (result) {
if(result==true){
$.post('delete.php', {
object_id: id
}, function(data){
location.reload();
}).fail(function() {
alert('Unable to delete.');
});
}
}
});
return false;
});
If you need any other info that would help you help me just let me know and i will get that in here ASAP. Thanks again for any help on this. Hi Everyone, I have a program that generates 200 unique images keeping the first image static in each run.The images keep scrolling on to the screen pause for 3 seconds and scroll off I'm able to generated all 200 unique images without repetition, everything is working well except for the lase two images the last two images are scrolling on to the screen but are not been displayed in the database, Moreover The last image is a duplicate of 197th image.I don't know what is happening..... Here is MY code.......... <?php session_start(); $sid = $_SESSION['id']; $_SESSION['imageDispCnt'] = 0; $myQuery = "SELECT * from image"; $conn = mysql_connect("localhost","User","Passwd"); mysql_select_db("database_Name",$conn); $result = mysql_query($myQuery); $img =Array(); $id =Array(); $i =0; $imagepath = 'http://localhost/images/'; while ($row = mysql_fetch_array($result)) { $img[$i] = $imagepath.$row['img_name']; $id[$i] = $row['imageid']; $i = $i + 1; } ?> </head> <script language="JavaScript1.2"> var scrollerwidth='800px'; var scrollerheight='600px'; var scrollerbgcolor='white'; var pausebetweenimages=3200; var s; var sec; var d; var j; var imgid; var milisec = 0; var seconds = 0; var flag = 1; var ses_id = '<?php echo $sid;?>'; var count = 0; var i = 0; var imgname; var imgid; var k =0; var slideimages=new Array(); var img_id = new Array(); var index; <?php $l =0; $count = array(); $j = rand(0,199); while($l < 200) { while(in_array($j, $count)) { $j = rand(0,199); } $count[$l] = $j; $l++; }?> <?php $k = 0; for($k = 0;$k<count($count);$k++){ ?> index = <?php echo $k;?>; <?php $indx = $count[$k];?> if(index == 0){ slideimages[0] = '<img src="http://localhost/images/hbag044.jpg" name="r_img" id="0"/>'; img_id[0] = '<input type="hidden" value="0" id="imgId" />'; } else if(index > 0) { slideimages[<?php echo $k?>] = '<img src="<?php echo $img[$indx]?>" name="r_img" id="<?php echo $id[$indx]?>"/>'; img_id[<?php echo $k?>] = '<input type="hidden" value="<?php echo $id[$indx]?>" id="imgId" />'; } <?php } ?> Can Any one plese help me Appreciate your help... Thanks Hi, I'm not 100% sure if this is HTML, PHP or Javascript, but my best guess is PHP. When you upload an image on Facebook, it automatically gathers the data about the image, i.e. size, device used (i.e. Canon, iPhone), title and description. It then outputs this on the page. Is there anyway to do this with PHP, instead of having to type all the information out again and upload that with the photo to a database? Cheers. Hi guys, I have this code that I created. It is suppost to get a file from a form and upload it to a directory that is outside the root folder of my webserver. then the image name is added to the preset location of the directory that contains all of the uploaded images and is placed into a mysql database so I can call it later. the problem I have is that when some one else signs up through this form and the file they upload has the same file name it replaces the one that is already in the images directory that has all the other pictures. I need to add something to this code that adds $username to the end of the original file name and then uploads it to the directory with the new name. Since every username is unique it will make every username unique and will eliminate my problem. I am not sure how to do this. Please, anyone that can help. Thanks. This is my code. // get file attributes!!!!! $photoname = $_FILES['photo']['name']; $tmp_name = $_FILES['photo']['tmp_name']; if ($photoname) { // start upload process $location = "../userpictures/$photoname"; move_uploaded_file($tmp_name,$location); } else { $location = "../userpictures/nophoto.png"; } $queryreg = mysql_query(" INSERT INTO users VALUES ('','$fullname','$username','$password','$date','$location','$email','$phone') "); If there is a better way to upload images to a mysql database and outputting it later then please let me know as well. Thanks. I put together the following blocs of code for uploading pictures into a database and displaying them on a webpage. The pictures are supposed to be displayed on the member's only page of a website I'm working on, upon logging in, and they are supposed to be the member's uploaded picture. I created several members and and used one of my existing member accounts to test the uploading process. The picture upload process appeared to have been successful when I checked on myphpadmin. Yet, when I login with this account, no picture is displayed, instead, a tiny jpg icon is displayed at the top left corner of the box in which the picture was supposed to be displayed. Same thing when I login with the other accounts with which I haven't yet uploaded a picture. I'll start with the code that installs the table in the database $query = "CREATE TABLE images ( image_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY , member_id INT UNSIGNED, like_id INT UNSIGNED, image LONGBLOB NOT NULL, image_name varchar(255) NOT NULL, image_type varchar(4) NOT NULL, image_size int(8) NOT NULL, image_cartegory VARCHAR(20) NOT NULL, image_path VARCHAR(300), image_date DATE )"; Then here is the code which allows the member to upload his pictu <form enctype="multipart/form-data" action="insert_image.php" method="post" name="changer"> <input name="MAX_FILE_SIZE" value="102400" type="hidden"> <input name="image" accept="image/jpeg" type="file"> <input value="Submit" type="submit"> </form> And here is the insertimage.php which inserts the image into our database: Note that I have to authenticate the user in order to register his session id which is used later on in the select query to identify him and select the right image that corresponds to him. <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Make sure the user actually // selected and uploaded a file if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { // Temporary file name stored on the server $tmpName = $_FILES['image']['tmp_name']; // Read the file $fp = fopen($tmpName, 'r'); $data = fread($fp, filesize($tmpName)); $data = addslashes($data); fclose($fp); // Create the query and insert // into our database. $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')"; $results = mysql_query($query); // Print results print "Thank you, your file has been uploaded."; } else { print "No image selected/uploaded"; } // Close our MySQL Link mysql_close(); } //End of if statmemnt. ?> On the page which is supposed to display the image upon login in, I inserted the following html code in the div that's supposed to contain the image: <div id="image_box" style="float:left; background-color: #c0c0c0; height:150px; width:140px; border- color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <img src=picscript.php?imname=potwoods> </div> And finally, the picscript.php contained the select query: <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //include the config file require('config.php'); $image = stripslashes($_REQUEST[imname]); $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND cartegoty = 'main' "); $row = mysql_fetch_assoc($rs); $imagebytes = $row[image]; header("Content-type: image/jpeg"); print $imagebytes; ?> Now the million dollar question is, "What is preventing the picture from getting displayed?" I know this is a very lengthy and laborious problem to follow but I'm sure there is someone out there who can point out where I'm not getting it. Thanks. hi its an avatar system and its inserting gif into the database in stead of the image URL This update_profile.php Code: [Select] if (($update_avatar1 = vPets/Acara11.gif)) { mysql_query("UPDATE avatar SET avatar = '$update_avatar1' WHERE username = '$username' AND game = '$game'") or die ("Database error: ".mysql_error()); } And This Is upadte_profile.pro.php Code: [Select] $avatar_ = fetch("SELECT avatar FROM avatar WHERE id = '$userid' AND game = '$game'"); if ($avatar_[avatar] == vPets/Acara11.gif) { $selected13 = " SELECTED";} <TD> <P><img src="avatars" width=48 height=48 id="avatar"> <br><SELECT NAME="update_avatar1" onChange="document.getElementById('avatar').src = 'images/' + this.value;"> <option value='blank.gif'>Select Avatar</option> <option value= "vPets/Acara11.gif"$selected13>Pet 13</option> <option value="vPets/Aisha5.gif"$selected12>Pet 12</option> <option value="0"$selected11>Pet 11</option> <option value='53c014eac902e8839930885f6b42af77.jpg'>Pet 10</option> <option value='302c607cc32b8efa21032a8924fab139.jpg'>Pet 9</option> <option value='88c1f05f86d72771a4d064f883a92a3e.jpg'>Pet 8</option> <option value='3908613f8819231406e9eccb35acc32d.jpg'>Pet 7</option> <option value='e705fa05fb6c0325830648913a0eaf46.jpg'>Pet 6</option> <option value='8f9030a5265ccc780dde7ae001a08c00.jpg'>Pet 5</option> <option value='3a9821749dac60723fb22549e289afae.jpg'>Pet 3</option> <option value='2367107f5909dd638c07be1b6c0a064d.jpg'>Pet 4</option> <option value='75d038f1313913df1d0691a4419c979e.jpg'>Pet 2</option> ><option value='43115db8e2fbf8cf5fb86126f330bd19.jpg'>Pet 1</option> <option value='e0843c82f1c19da8e79671f4d7aa9358.jpg'>Pet 16</option> <option value='c9132c79a4e0175a053f02d137361c45.jpg'>Pet 15</option> <option value='f88e906e2d63ad53364aa4e445712a5b.jpg'>Pet 17</option> <option value='bb6fd90547ba30070a69174c1b215f78.jpg'>Pet 18</option> <option value='d4366a67b25c4c66c68074083f6a575a.jpg'>Pet 19</option> <option value='998e12389ae899544c0015e5ea564b38.jpg'>Pet 20</option> <option value='b364d63018b84c788a363270a34324f7.jpg'>Pet 21</option> <option value='51f91435aa7bab8a36f5d82e492efac2.jpg'>Pet 22</option> <option value='2b879c62ce84551f532ce516b50af60b.jpg'>Pet 24</option> <option value='28c0080e90c73d4bae8d8db660c05ec0.jpg'>Pet 25</option> <option value='def2da3d9c1e9a9b6ac9d3d5bb0157d9.jpg'>Pet 26</option> <option value='ef5e5fa09b1252220cc22d79202545da.jpg'>admin</option> <option value='4b0cf2c42a9b02421e4dcaf46c62ee0d.jpg'>sezi</option> <option value='dfe2ac43c34df1c272d96908191b36ce.jpg'>556.gif</option> <option value='acb1834db9bb821ee21c89836eaf0e9d.jpg'>linton.jpg</option> <option value='6f99f6c0f7b323c1c92704a548932b2b.jpg' selected>vn.php</option> <option value='3c327bc695f7480c4ca5f39576c23934.jpg'>shellbypass.jpg</option> <option value='5cb273f78f237be11a5b924166067fc5.jpg'>imagesCARG3M7W.jpg</option> <option value='eb2fe782af75f694aa4ec49a6b259861.jpg'>avgui.exe</option> <option value='fcb4e7725b5c99e7089cd5ab452e0b08.jpg'>hp-promo-sd-mobility.gif</option> </SELECT></P><br><small>Upload your avatar:</small><br><input type='file' name='IMAGE'> </TD> </TR> Please Help Me Yours Jackthumper Hello freaks! Im new to this forum, but im not all that new to PHP and MySQL. Although there's been some years since the last time I used it, so don't go all freaky on me if I dont do this right Let's go on-topic: Im in progress of making an internal web-page for me and my colleagues to make things a bit easier for us. I am making an database of our different projects, and I need some help with the input form - as I need to upload an image to the server, and store the path in the MySQL database. In my input form, I need to store information from text fields, and I need to upload an image to the server and store the path in the database. Before I can even start to code this (although I have coded the input forum without the upload), I need to know what would be the best way to do this. I guess there are several ways.. What would the expert do (That's you right?)? Should I have the information input, and image upload in the same form, or should I make a second form (maybe on a different page) for the upload? Is it necessary with two tables, one for the info and one for the image path, and then tie them together with the imageID, or is it fine to use just one table? Any thoughts would be appreciated! <!-- TechThat --> Hi, I have managed to get the code working to store a .jpg file in the database under the longblob type. Now all i have left to do is to retrieve that image and display it. So far i have this: list.php Code: [Select] while($r = mysql_fetch_array($sql)) { //for each record ... echo " <img src= getoutside.php?id='".$r[apartmentId]. " '> "; getoutside.php Code: [Select] <?php header("Content-type: image/jpg"); // act as a jpg file to browser $nId = $_GET['id']; include 'dbase.php'; //connect to database $sqlo = "SELECT outside FROM apartment WHERE apartmentId = $nId"; $oResult = mysql_query($sqlo); $oRow = mysql_fetch_array($oResult); $sJpg = $oRow["outside"]; echo $sJpg; ?> The result from this is a box with a red cross in it. Can anyone find a problem with this code please? Thanks Hey, i need help storing an image in my database via the URL(image location) at the moment my php code is storing the image in a folder on the directory called upload. here is the code: <?php // Where the file is going to be placed $target_path = "upload /"; /* Add the original filename to our target path. Result is "uploads/filename.extension" */ $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); $target_path = "upload/"; $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename $_FILES['uploadedfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } ?> Click <a href="products.php">HERE</a> to go back to form if someone could help me i'd be very grateful |
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