PHP - Show Image From Mysql In Php
Hi guys, I have page where it echos out the image url from mysql in a MAMP Server,
however when i echo the image url out it seems to be fine but as soon as i put in a img scr it wont show the image but it shows the container. I have the code here Code: [Select] <div id="maincontentholderbottom"> <div id="maincontentholderbottom-index-left"> <div id="news-container"><ul> <?php $select=mysql_query("SELECT * FROM news"); while($get_news=mysql_fetch_array($select)){ $newsid=$get_news['id']; $title=$get_news['title']; $text=$get_news['text']; $newsdate=$get_news['date']; $newstime=$get_news['time']; $imagelink=$get_news['newsimagelink']; $newstext=substr($text, 0, 420); echo"<li><div id='title'>$title</div><div id='newsblock'>$newstext... <a href='http://localhost/mycomputer/create/news/index.php?id=$newsid'>Read More</a></div> <div id='newsimage'><img src='$imagelink' width='150' height='70'/></div> </li>"; } ?> </ul> </div> </div> </div> Do u know why is it like this? I appreciate your help in advance. Thanks! Similar Tutorialscreate table mimi (mimiId int(11) not null, mimiBody varchar(255) ); <?php //connecting to database include_once ('conn.php'); $sql ="SELECT mimiId, mimiBody FROM mimi"; $result = mysqli_query($conn, $sql ); $mimi = mysqli_fetch_assoc($result); $mimiId ='<span>No: '.$mimi['mimiId'].'</span>'; $mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>'; ?> //what is next? i want to download pdf or text document after clicking button or link how to do that Hi, I have some images that I want to resize to a square dimention (40px 40px). However, not all image are square, so when they are resized, they lose their ration and look squashed. Is there a way I can display just a part of the original image, say 40px x 40px on the top middle of each image? I hope that makes sense Hi all I have created a mySQL query and assigned it to a string: Code: [Select] $getjobs = mysql_query (" SELECT * FROM jobs WHERE from_date >= '".substr($newdate, 0, 10)."' $keyword_search $sector_search $salary_search $location_search $type_search ORDER BY job_refno DESC ") or die (mysql_error()); Is there any way to output the actual query to the browser to check it? Hi all, I am trying to make a members details section that can be updated. I want to be able to "SELECT * FROM users WHERE email='$email'" and then show the values that can be changed in a html drop down box with the selection that was made when the user registered already selected="selected"; You will be able to see what I am attempting to do below. <?php $sql = "SELECT manufacturer FROM table1 WHERE email=$'email'"; $result = mysqli_query($cxn,$sql); $row = mysqli_fetch_assoc($result); foreach manufacturer in table1 { if table1.manufacturer = table2.manufacturer { echo '<option name="manufacturer" selected="selected" value"$row['manufacturer']"</option>'; } else { echo '<option name="manufacturer" value"$row['manufacturer']"</option>'; } } ?> Hi all expert. I am a newbie in this PHP programming. I need your help or advise on the PHP. And question is, I have a list of data and the details are as below:
ID BILLNO DATE AMOUNT ITEM DESCRIPTION QTY UPRICE 1 IV001 01/01/2015 100.00 A1 Balloon 1 30.00 2 IV001 01/01/2015 100.00 A2 Bag 2 20.00 3 IV001 01/01/2015 100.00 A3 Pen 3 10.00 4 IV002 02/01/2015 20.00 A3 Pen 2 10.00 5 IV003 02/01/2015 50.00 A1 Balloon 1 30.00 6 IV003 02/01/2015 50.00 A2 Bag 1 20.00 How can I make the output in xml by using PHP to output as below: <RECORD> <HEADER BILLNO="IV001" DATE="01/01/2015" AMOUNT="100.00> <DETAIL ITEM="A1" DESCRIPTION="Balloon" QTY="1" UPRICE="30.00"> </DETAIL> <DETAIL ITEM="A2" DESCRIPTION="Bag" QTY="2" UPRICE="20.00"> </DETAIL> <DETAIL ITEM="A3" DESCRIPTION="Pen" QTY="3" UPRICE="10.00"> </DETAIL> </HEADER> <HEADER BILLNO="IV002" DATE="02/01/2015" AMOUNT="20.00> <DETAIL ITEM="A3" DESCRIPTION="Balloon" QTY="2" UPRICE="10.00"> </DETAIL> </HEADER> </RECORD> Your feedback is highly appreciated. Thank you. Hi again PhpFreaks, yet again I got a problem regarding coding. I currently have php code that outputs information from my database, but how do i make it also display the name of the fields beside each entry of the database? heres my PHP code: Code: [Select] <?php $server = ""; // Enter your MYSQL server name/address between quotes $username = ""; // Your MYSQL username between quotes $password = ""; // Your MYSQL password between quotes $database = ""; // Your MYSQL database between quotes $con = mysql_connect($server, $username, $password); // Connect to the database if(!$con) { die('Could not connect: ' . mysql_error()); } // If connection failed, stop and display error mysql_select_db($database, $con); // Select database to use // Query database $result = mysql_query("SELECT * FROM Properties"); if (!$result) { echo "Error running query:<br>"; trigger_error(mysql_error()); } elseif(!mysql_num_rows($result)) { // no records found by query. echo "No records found"; } else { $i = 0; echo '<div style="font-family:helvetica; font-size:15px; padding-left:15px; padding-top:20px;">'; while($row = mysql_fetch_array($result)) { // Loop through results $i++; echo '<img class="image1" src="'. $row['images'] .'" />'; //image echo "Displaying record $i<br>\n"; echo "<b>" . $row['id'] . "</b><br>\n"; // Where 'id' is the column/field title in the database echo $row['Location'] . "<br>\n"; // Where 'location' is the column/field title in the database echo $row['Property_type'] . "<br>\n"; // as above echo $row['Number_of_bedrooms'] . "<br>\n"; // .. echo $row['Purchase_type'] . "<br>\n"; // .. echo $row['Price_range'] . "<br>\n"; // .. } echo '</div>'; } mysql_close($con); // Close the connection to the database after results, not before. ?> thanks in advance I'm trying to show the same data, but updated right away. For example. I want to update my coords on a map and refresh a div to show the new data, but as the code now, it keeps the same data until I reload the page. Here is the code I have now. if ($north) { $ylocation = $users['y'] + 1; if ($ylocation > 5) { $ylocation = 0; } $locationyupdate = ("UPDATE players SET y = '$ylocation' WHERE name='$users[name]'"); mysql_query($locationyupdate) or die("could not register");?> <script type="text/javascript"> $('#npc').load('npc.php'); $('#description').load('description.php'); </script><?} The update code is before the reload script for the two div's. The data DOES change in the database, but the two div's won't display the new data until it is refreshed again. Do I need to reactivate fetch to get the new data? I have a commenting system and i have a limit of a certain number of comments to be shown. What i want to do is have a button on the bottom of the page at the end of the comments that are showing and when you click it ajax loads the next certain number of of rows (but not all of them),and then you click it again and it shows more of them, etc. So for example. comment 1 comment 2 comment 3 comment 4 --click button--(loads 4 more)--- comment 5 comment 6 comment 7 comment 8 --click button--(loads 4 more)-- comment 9 comment 10 comment 11 comment 12 etc. until there are no more rows. what's the best way to do this? (I know how to do the ajax and all, i just need help with the script to select the rows) Thanks. Hi, my page is currently up at brewhas.org/transactions.php. When I retrieve a set of results > the defined # per page (such as entering 'WI' for last name), I see the first page correctly, and see the links to next pages, but when I go to the next page there are no results displayed. Same for p.3, etc. So I've looked around enough to know that I need to pass a variable to the subsequent pages, either via a session variable or the URL. I have 2 fields input on the form, as well as a limit and offset. My questions are 1) do I need to pass all 4 and if so then should I use a session variable so the URL does not get too long, and 2) can anyone provide any help with setting session variables and including them to be passed? I'm a newbie and have grabbed some code where I can but this is where i'm stuck. thanks in advance. Code: [Select] <?php include 'config.php'; include 'opendb.php'; $rows_per_page = 20; //Get the values from the text boxes $fnamesearch = $_POST["fnamesearch"]; $lnamesearch = $_POST["lnamesearch"]; //These trim and convert the name fields to upper case. $fnamesearch=strtoupper($fnamesearch); $lnamesearch=strtoupper($lnamesearch); $fnamesearch=trim($fnamesearch); $lnamesearch=trim($lnamesearch); // Count how many rows are coming back for the query //This checks to see if there are at least 2 chars in the last name. if (strlen($lnamesearch)<2) { echo "<p class='style7'>Please enter at least 2 letters of the last name.</p>"; exit; } else //sets the query to get the number of rows { $query = "SELECT COUNT(*) FROM BKL_TRANSACTIONS WHERE((PLAYER_LNAME LIKE '$lnamesearch%')& (PLAYER_FNAME LIKE '$fnamesearch%'))"; } $result = mysql_query($query) or die ("Could not execute the query"); //executes the get count query $query_data = mysql_fetch_row($result); $numrows = $query_data[0]; echo "count: ".$numrows; $lastpage = ceil($numrows/$rows_per_page); //determines how many pages based on rows divided by rows per page echo "pages: ".$lastpage; // Get required page number. If not present, default to 1. if (isset($_GET['pageno']) && is_numeric($_GET['pageno'])) {$pageno = $_GET['pageno']; } else {$pageno = 1; } //Checks that the value of $pageno is an integer between 1 and $lastpage. $pageno = (int)$pageno; if ($pageno > $lastpage) {$pageno = $lastpage; } if ($pageno < 1) {$pageno = 1; } //constructs OFFSET for the sql SELECT statement $offset = ($pageno - 1) * $rows_per_page; //This checks to see if there are at least 2 chars in the last name. if (strlen($lnamesearch)<2) { echo "<p class='style7'>Please enter at least 2 letters of the last name.</p>"; exit; } else //sets the query to get the number of rows { $query = "SELECT * FROM BKL_TRANSACTIONS WHERE((PLAYER_LNAME LIKE '$lnamesearch%')& (PLAYER_FNAME LIKE '$fnamesearch%')) LIMIT $offset, $rows_per_page"; } $result = mysql_query($query) or die ("Could not execute the query"); //executes the query //Count the number of results $numrows=mysql_num_rows($result); echo "<p class='style7'>Your search: "".$fnamesearch." ".$lnamesearch."" returned <b>".$numrows."</b> results.</p>"; if ($numrows == 0) //if no matches, don't display the table exit; else //build the table and insert rows { echo "<table border=1 cellspacing=0 cellpadding=1 width='77%' align=left bordercolor=#666666>"; echo "<tr bgcolor=#cccc99 class='style5'><th width='10%' class='style5'>Date</th>"; echo "<th width='10%' class='style5'>Action</th>"; echo "<th width='12%' class='style5'>From</th>"; echo "<th width='12%' class='style5'>To</th>"; echo "<th width='5%' class='style5'>Pos</th>"; echo "<th width='18%' class='style5'>Name</th>"; echo "<th width='5%' class='style5'>Round</th></tr>"; while ($row = mysql_fetch_array($result)) { echo "<tr><td width='10%' class='style6'>".$row['TRANS_DT']."</td>"; echo "<td width='10%' class='style6'>".$row['TRANS_ACTION']."</td>"; echo "<td width='12%' class='style6'>".$row['FROM_OWNER']."</td>"; echo "<td width='12%' class='style6'>".$row['TO_OWNER']."</td>"; echo "<td width='5%' class='style6'>".$row['POSITION']."</td>"; echo "<td width='18%' class='style6'>".$row['PLAYER_FNAME']." ".$row['PLAYER_LNAME']."</td>"; echo "<td width='5%' class='style6'>".$row['DRAFT_RND']."</td></tr>"; } /****** build the pagination links ******/ // if not on page 1, don't show back links if ($pageno > 1) { // show << link to go back to page 1 echo " <a href='{$_SERVER['PHP_SELF']}?pageno=1'><<</a> "; // get previous page num $prevpage = $pageno - 1; // show < link to go back to 1 page echo " <a href='{$_SERVER['PHP_SELF']}?pageno=$prevpage'><</a> "; } // range of num links to show $range = 3; // loop to show links to range of pages around current page for ($x = ($pageno - $range); $x < (($pageno + $range) + 1); $x++) { // if it's a valid page number... if (($x > 0) && ($x <= $lastpage)) { // if we're on current page... if ($x == $pageno) { // 'highlight' it but don't make a link echo " [<b>$x</b>] "; // if not current page... } else { // make it a link echo " <a href='{$_SERVER['PHP_SELF']}?pageno=$x'>$x</a> "; } // end else } // end if } // if not on last page, show forward and last page links if ($pageno != $lastpage) { // get next page $nextpage = $pageno + 1; // echo forward link for next page echo " <a href='{$_SERVER['PHP_SELF']}?pageno=$nextpage'>></a> "; // echo forward link for lastpage echo " <a href='{$_SERVER['PHP_SELF']}?pageno=$lastpage'>>></a> "; } // end if /****** end build pagination links ******/ echo "</table>"; } mysql_free_result($result); //release the result set from the table mysql_close($conn); //close the connection to the db ?> MOD EDIT: [code] . . . [/code] tags added. Hi, How can I get others informations on another page by clicking on one row's value I have this code, but it doesn't work: while($row = mysql_fetch_array($result)) { echo "<tr><p>"; echo "<th><p>" . $row['fname'] . " " . $row['lname'] . "</p></th>"; echo"<th><p><a href='student.php?id='".$row['topic']."'\'>".$row['topic']."</p></a></th>"; echo "</tr>"; } echo "</table>"; Student.php <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("college", $con); $result = mysql_query("SELECT * FROM students"); $topic = $_Get['topic']; while($row = mysql_fetch_array($result)) { echo "<tr><p>"; echo "<th><p>" . $row['month']." " .$row['datetime']."</p></th>"; echo"<th><p>". $row['topic'] ."</p></th>"; echo "<th><p>" . $row['gender'] . "</p></th>"; echo "<th><p>" . $row['fname'] . " " . $row['lname'] . "</p></th>"; echo "<th ><p>" . $row['id'] . "</p></th>"; echo "</tr>"; echo "</table>"; mysql_close($con); } ?> Hey guys i have a query echo (empty($row['deal'])? "empty": "not empty"); //result not empty but i need it to show the "not empty" if its "deal". by default it is a "1" when i select the row to be a deal it updates the table and inserts "Deal". Hi again probably a very simple code but not working. I am trying to show a default image "fabric.jpg" if the recordset is empty. If not empty it shows the recordset image at a size of 100X100. This is the code I am using and have probably left something out, any ideas? <img src="<?php if ($row_Recordset3['fabricpicture']==null); echo "<img src='graphics/fabric.jpg'>"; ?>" alt="" name="fabric" width="100" height="100" border="0" align="bottom" id="fabric" title="Selected Bottom Up Blind Fabric" /> I have tried switching the img scr both "" and ' ' but still no joy? Hey everyone, I could use some help with this one.. I need to create some php code to switch images when a certain page loads. For example: These are the pages, test1, test2,test3, so when you click on the test1 link in the navigation a certain image would show up : something like this: If test1 exists then show image1.jpg If test2 exists then show image2.jpg If test3 exists then show image2.jpg. I am doing this in wordpress and I can't figure out the if then statement.. Does anyone have any ideas?? thanks Hello I am having problems uploading an image through a HTML form. I want the image to be uploaded to the server and the image name to be written to the mysql database. Below is the code I am using: Code: [Select] <?php if (isset($_POST['add'])){ echo "<br /> add value is true"; $name = $_POST['name']; $description = $_POST['description']; $price = $_POST['price']; $category_id = $_POST['category_name']; $image = $_FILES['image']['name']; //file path of the image upload $filepath = "../images/"; //mew name for the image upload $newimagename = $name; //new width for the image $newwidth = 100; //new height for the image $newheight = 100; include('../includes/image-upload.php'); mysql_query("INSERT INTO item (item_name, item_description, item_price, item_image) VALUES ('$name','$description','$price','$image')"); ?> Here is the image-upload.php file code: Code: [Select] <?php //assigns the file to the image $image =$_FILES["image"]["name"]; $uploadedfile =$_FILES["image"]["tmp_name"]; if ($image) { //retrieves the extension type from image upload $extension = getextension($image); //converts extension to lowercase $extension = strtolower($extension); //create image from uploaded file type if($extension=="jpg" || $extension=="jpeg") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefromjpeg($uploadedfile); }else if($extension=="png") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefrompng($uploadedfile); }else{ $src = imagecreatefromgif($uploadedfile); } //creates a list of the width and height of the image list($width,$height)=getimagesize($uploadedfile); //adds color to the image $tmp = imagecreatetruecolor($newwidth,$newheight); //create image imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height); //set file name $filename = $filepath.$newimagename.".".$extension; $imagename = $newimagename.".".$extension; //uploads new file with name to the chosen directory imagejpeg($tmp,$filename,100); //empty variables imagedestroy($src); imagedestroy($tmp); } ?> Any help would be appreciated, fairly new to all this! Thanks!!! I need to be able to create an image from user input. I have found a couple of samples that purport to do that but when I copy the code I can't get it to work. I would appreciate advice as to what I am doing wrong please. One example I tried is from a tutorial on this site: http://www.phpfreaks.com/tutorial/php-add-text-to-image Now perhaps there is something I am doing wrong when I call on the image that is supposed to display but given that I have tried the same code with other images and it works I can't see what I am doing wrong, unless there is something in the nature of scripts that requires some extra steps. Here is my code to see the image (note that 'myscript.php' in my code is in the same folder as the code I am calling: <html> <head> <title>Using Images Created by Scripts</title> </head> <body> <h1>Generated Image Below ...</h1> <img src="myscript.php"/> </body> </html> Can anyone help? Thank you I want to show a cookie of a referral's username on a sign up page. The link is like this, www.mysite.com/signup?ref=johnsmith. The cookie doesn't show if I go to that url page. But it does show up once I reload the page. So I'm wondering if it's possible to show the cookie the first time around, instead of reloading the page? Here is my code. // This is in the header $url_ref_name = (!empty($_GET['ref']) ? $_GET['ref'] : null); if(!empty($url_ref_name)) { $number_of_days = 365; $date_of_expiry = time() + 60 * 60 * 24 * $number_of_days; setcookie( "ref", $url_ref_name, $date_of_expiry,"/"); } else if(empty($url_ref_name)) { if(isset($_COOKIE['ref'])) { $user_cookie = $_COOKIE['ref']; } } else {} // This is for the sign up form if(isset($_COOKIE['ref'])) { $user_cookie = $_COOKIE['ref']; ?> <fieldset> <label>Referred By</label> <div id="ref-one"><span><?php if(!empty($user_cookie)){echo $user_cookie;} ?></span></div> <input type="hidden" name="ref" value="<?php if(!empty($user_cookie)){echo $user_cookie;} ?>" maxlength="20" placeholder="Referrer's username" readonly onfocus="this.removeAttribute('readonly');" /> </fieldset> <?php }
hello experts ! im new to php script, hope u guys able to help. thanks in advance . I'm following a tutorial from this site: http://www.anyexample.com/programming/php/php_mysql_example__image_gallery_(blob_storage).xml and i run the code, error occur (i highlighted line 27) : Parse error: syntax error, unexpected T_STRING in C:\xampp\htdocs\testing.php on line 27 This is my php script: <?php $db_host = 'localhost'; // don't forget to change $db_user = 'root'; $db_pwd = '123456'; $database = 'test'; $table = 'ae_gallery'; // use the same name as SQL table $password = '123'; // simple upload restriction, // to disallow uploading to everyone ////create sql-table CREATE TABLE 'ae_gallery' ( 'id' int(11) NOT NULL AUTO_INCREMENT, 'title' varchar(64) character SET utf8 NOT NULL, 'ext' varchar( 8 ) character SET utf8 NOT NULL, 'image_time' timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP, 'data' mediumblob NOT NULL, PRIMARY KEY (`id`) ); if (!mysql_connect($db_host, $db_user, $db_pwd)) die("Can't connect to database"); if (!mysql_select_db($ae_gallery)) die("Can't select database"); // This function makes usage of // $_GET, $_POST, etc... variables // completly safe in SQL queries function sql_safe($s) { if (get_magic_quotes_gpc()) $s = stripslashes($s); return mysql_real_escape_string($s); } // If user pressed submit in one of the forms if ($_SERVER['REQUEST_METHOD'] == 'POST') { // cleaning title field $title = trim(sql_safe($_POST['title'])); if ($title == '') // if title is not set $title = '(empty title)';// use (empty title) string if ($_POST['password'] != $password) // cheking passwors $msg = 'Error: wrong upload password'; else { if (isset($_FILES['photo'])) { @list(, , $imtype, ) = getimagesize($_FILES['photo']['tmp_name']); // Get image type. // We use @ to omit errors if ($imtype == 3) // cheking image type $ext="png"; // to use it later in HTTP headers elseif ($imtype == 2) $ext="jpeg"; elseif ($imtype == 1) $ext="gif"; else $msg = 'Error: unknown file format'; if (!isset($msg)) // If there was no error { $data = file_get_contents($_FILES['photo']['tmp_name']); $data = mysql_real_escape_string($data); // Preparing data to be used in MySQL query mysql_query("INSERT INTO {$table} SET ext='$ext', title='$title', data='$data'"); $msg = 'Success: image uploaded'; } } elseif (isset($_GET['title'])) // isset(..title) needed $msg = 'Error: file not loaded';// to make sure we've using // upload form, not form // for deletion if (isset($_POST['del'])) // If used selected some photo to delete { // in 'uploaded images form'; $id = intval($_POST['del']); mysql_query("DELETE FROM {$table} WHERE id=$id"); $msg = 'Photo deleted'; } } } elseif (isset($_GET['show'])) { $id = intval($_GET['show']); $result = mysql_query("SELECT ext, UNIX_TIMESTAMP(image_time), data FROM {$table} WHERE id=$id LIMIT 1"); if (mysql_num_rows($result) == 0) die('no image'); list($ext, $image_time, $data) = mysql_fetch_row($result); $send_304 = false; if (php_sapi_name() == 'apache') { // if our web server is apache // we get check HTTP // If-Modified-Since header // and do not send image // if there is a cached version $ar = apache_request_headers(); if (isset($ar['If-Modified-Since']) && // If-Modified-Since should exists ($ar['If-Modified-Since'] != '') && // not empty (strtotime($ar['If-Modified-Since']) >= $image_time)) // and grater than $send_304 = true; // image_time } if ($send_304) { // Sending 304 response to browser // "Browser, your cached version of image is OK // we're not sending anything new to you" header('Last-Modified: '.gmdate('D, d M Y H:i:s', $ts).' GMT', true, 304); exit(); // bye-bye } // outputing Last-Modified header header('Last-Modified: '.gmdate('D, d M Y H:i:s', $image_time).' GMT', true, 200); // Set expiration time +1 year // We do not have any photo re-uploading // so, browser may cache this photo for quite a long time header('Expires: '.gmdate('D, d M Y H:i:s', $image_time + 86400*365).' GMT', true, 200); // outputing HTTP headers header('Content-Length: '.strlen($data)); header("Content-type: image/{$ext}"); // outputing image echo $data; exit(); } ?> <html><head> <title>MySQL Image Gallery Example</title> </head> <body> <?php if (isset($msg)) // this is special section for // outputing message { ?> <p style="font-weight: bold;"><?=$msg?> <br> <a href="<?=$PHP_SELF?>">reload page</a> <!-- I've added reloading link, because refreshing POST queries is not good idea --> </p> <?php } ?> <h1><i>Image gallery</i></h1> <h2>Uploaded images:</h2> <form action="<?=$PHP_SELF?>" method="post"> <!-- This form is used for image deletion --> <?php $result = mysql_query("SELECT id, image_time, title FROM {$table} ORDER BY id DESC"); if (mysql_num_rows($result) == 0) // table is empty echo '<ul><li>No images loaded</li></ul>'; else { echo '<ul>'; while(list($id, $image_time, $title) = mysql_fetch_row($result)) { // outputing list echo "<li><input type='radio' name='del' value='{$id}'>"; echo "<a href='{$PHP_SELF}?show={$id}'>{$title}</a> – "; echo "<small>{$image_time}</small></li>"; } echo '</ul>'; echo '<label for="password">Password:</label><br>'; echo '<input type="password" name="password" id="password"><br><br>'; echo '<input type="submit" value="Delete selected">'; } ?> </form> <h2>Upload new image:</h2> <form action="<?=$PHP_SELF?>" method="POST" enctype="multipart/form-data"> <label for="title">Title:</label><br> <input type="text" name="title" id="title" size="64"><br><br> <label for="photo">Photo:</label><br> <input type="file" name="photo" id="photo"><br><br> <label for="password">Password:</label><br> <input type="password" name="password" id="password"><br><br> <input type="submit" value="upload"> </form> hello world! First post here! I'm a newbie trying to make my own cms, everything is working except the import of image into the post, this becomes text. Here is the code: Code: [Select] function connect() { $con = mysql_connect($this->host, $this->username, $this->password) or die (mysql_error()); mysql_select_db ($this->db, $con) or die (mysql_error()); } function get_content($id = '') { if($id != ""): $id = mysql_real_escape_string($id); $sql = "SELECT * FROM cms_content WHERE id = '$id'"; else: $sql = "SELECT * FROM cms_content ORDER BY id DESC"; endif; $return = '<a href="index.php">go back</a>'; $res = mysql_query($sql) or die(mysql_error()); if(mysql_num_rows($res) != 0): while($row = mysql_fetch_assoc($res)) { echo '<div class="title"><h1><a href="index.php?id=' . $row['id'] . '">' . $row['title'] . '</a></h1></div>'; echo '<div class="date"><p>' . $row['date'] . '</p></div>'; echo '<div class="photo"><p>' . $row['photo'] . '</p></div>'; echo '<div class="txt"><p>' . $row['body'] . '</p></div>'; } else: echo '<p class="results_">uH oh!</p>'; endif; echo $return; } function add_content($p) { $title = mysql_real_escape_string($p['title']); $date = mysql_real_escape_string($p['date']); $photo = mysql_real_escape_string($p['photo']); $body = mysql_real_escape_string($p['body']); // if they write wrong if(!$title | | !$body): if(!$title): echo "<p>The Title is Required</p>"; endif; if(!$photo): echo "<p>No photo</p>"; endif; if(!$body): echo "<p>The body is Required</p>"; endif; echo '<p><a href="add-content.php">Try again</a></p>'; else: // add to db $sql = "INSERT INTO cms_content VALUES (null, '$title', '$date', '$photo', '$body')"; $res = mysql_query($sql) or die(mysql_error()); echo "<p class='results_'>Added with success!</p>"; // [END] add to db endif; // [END] if they write wrong } and in the html i got a form Code: [Select] <form enctype="multipart/form-data" method="post" action="index.php"> <input type="hidden" name="add" value="true"> <div> <label for="title"><p>Title:</p></label> <input type="text" name="title" id="title" /> </div> <div> <input type="file" name="photo"><br> </div> <div> <label for="body"><p>Body:</p></label> <textarea name="body" id="body" rows="8" cols="40"></textarea> </div> <input type="submit" name="submit" value="Add Content"> </form> |