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PHP - Display Mediumblob Image Help
it's not displaying the image... any help?
Code: [Select] echo "<img src='advertimage.php?id=$advertid' />"; advertimage.php Code: [Select] <?php $id= $_GET['id']; $thedatabase = new Database2(); $thedatabase->opendb2(); $advert = $thedatabase->viewadvert($id); header("Content-type: image/jpeg"); echo $advert; $thedatabase->closedb2(); ?> viewadvert function Code: [Select] function viewadvert($id){ $getadvert = mysql_query("SELECT * FROM `advertise` WHERE `id`='$id' LIMIT 1",$this->connect); $row=mysql_fetch_assoc($getadvert); $image = $row['image']; return $image; } Similar TutorialsWould like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance. BTW, is using php only doable? Is there a simpler or more elegant way to do this? Thanks all! 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. I have a script that uploads my images to a folder and database which works fine. I am looking to display my images on the fly to resize them. How do I intergrate my image output string with the fly string. Current output Code: [Select] echo '<img src="' . $dir . '/' . $image_id . '.jpg">'; Fly string Code: [Select] <img src="/scripts/thumb.php?src=/images/whatever.jpg&h=150&w=150&zc=1" alt="" /> Thanks for your help Hey there I am having a problem where I am pulling the records for images from my database and then trying to display them. What happens is there are three images in the database and I want to display them. My code displays the images but it only shows two of the three. I know there are three images in the database I put them in and checked just to make sure. It just seems to only want to display two images instead of three. Here's the code:$getImages = "SELECT * FROM Images WHERE SKU = '".$sku."'"; $resultGetImages = mysql_query($getImages); $records = mysql_fetch_assoc($resultGetImages); ?> while($records = mysql_fetch_assoc($resultGetImages)){ echo "<a onclick=\"document.getElementById('placeholder').src='../../".$records['Path']."'\" target=\"_blank\"><img src=\"../../".$records['ThumbPath']."\" alt=\"".$records['Description']."\" height=\"".$records['ThumbWidth']."\" /></a>\n"; } ?> I have tried a for loop as well but it displayed more images than were in the database and just a single image over and over again. I know that was a flaw in my code and I'm not even sure a for loop would even be a reasonable way to go. Don't know if it is php or html but the image won't display. Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en" > <head> <meta name="google-site-verification" content="y_kspZkCZOQM-WGNB_lQ8BhJS8p7-B66hkHI_UhzKbs" /> <meta http-equiv="Content-type" content="text/html; charset=utf-8" /> <meta http-equiv="Content-language" content="en" /> <link type="text/css" rel="stylesheet" href="css/reset.css" /> <link type="text/css" rel="stylesheet" href="css/960.css" /> <link type="text/css" rel="stylesheet" href="css/custom.css" /> <title>Bay Area Remote Control Society</title> </head> <body> <div id="wrapper" class="container_12"> <div id="header" class="grid_12"> <div id="left-header" class="grid_5 alpha"><?php include('content/left-header.php'); ?></div> <div id="newsflash" class="grid_7 omega"><?php include('content/newsflash.php'); ?></div> </div> <div id="leftmenu" class="grid_3"><?php include('content/menu2.php'); ?></div> <div id="maincontent" class="grid_9"><?php include('content/maincontent.php'); ?></div> <div id ="footer" class="grid_12"><?php include('content/footer.php'); ?></div> </div><!-- end wrapper --> </body> </html> left-header.php Code: [Select] <?php ?> <img src="Pictures_Other/concorde.jpg" alt="concorde" border="0" width="250" height="188" /> Hi im wanting to create this system where it loads all your files in a specified directory, loads all the images, although i have done that what im trying to do is have it so it prints them into a 9x9 grid automatically, in which i need to find something to restrain the echo of the variable to a limit of 9.. i need to create the <tr><th>and <td>'s automatically and also id like it so if there is more than 9, it will allow me to select a button which will load up the next section of images. heres the code i have so far: <?php $dir = "images/"; $dh = opendir($dir); while (false !== ($filename = readdir($dh))) { $files[] = $filename; } sort($files); print_r($files); I then used this to try and echo into a table :S <?php if ($handle = opendir('images/')) { while (false !== ($file = readdir($handle))) { echo "<table width='450px' height'450px' border='4px solid #000'>"; if ($file != "." && $file != ".."){ echo "<tr><td>$file<br></td>"; echo "<td><img src=images/".$file." height='20px' width='20px'></td></tr><br>"; } } echo "</table>"; closedir($handle); } ?> They are seperate as i havent merged together as i wud like to get them working first. I was trying to do the max and min on the array numbers but was stuck doing that. Any help would be greatly apprecated Many thanks Lewis Stevens Hey there, my problem is that I am trying to pull three images from a database and then displaying them with a large image above them. I want to then be able to click the smaller ones and then they become the larger one(kind of like on amazon). I am having 2 major problems with it at this point. the first is my while statement is only pulling 2 of the three images from the database, and secondly my image replacement javascript replaces with the smaller image. With the javascript it dosen't have to be done that way or in javascript so a link or suggestions on how to do that better would be great. So here's my code: <?php include('../../php/includes/openDbConn.php'); $sku = 'c-111111'; $getImages = "SELECT * FROM Images WHERE SKU = '".$sku."'"; $resultGetImages = mysql_query($getImages); $records = mysql_fetch_assoc($resultGetImages); ?> <img src="../../<?php echo $records['Path'] ?>" id="placeholder" alt="<?php echo $record['Description']; ?>" height="<?php echo $record['Width']; ?>" /> <div id="thumbImages"> <?php while($records = mysql_fetch_assoc($resultGetImages)){ echo "<a onclick=\"document.getElementById('placeholder').src='../../".$records['ThumbPath']."'\" target=\"_blank\"><img src=\"../../".$records['ThumbPath']."\" alt=\"".$records['Description']."\" height=\"".$records['ThumbWidth']."\" /></a>\n"; } ?> and a bit on the table with the images. It stores the sku and the Path of a thumbnail, regular and large sized image. All help is much appreciated. Hi, Im trying to get an image to display from a reference in the database but am having no joy. At the minute Im just playing about with PHP to get familiar with it so you can ignore most of the code. The line im interested in is: <td><?php echo "<img src=\"C:\wamp\www\fermpix\Pics\'{$row["Name"]}'\">";?></td> When I view the page in my browser I see the attached. Can anybody see what im doing wrong? Heres the full page: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <title>Browse Upload Files</title> </head> <body bgcolor="white"> <?php error_reporting(E_ALL); include 'db.inc'; $query = "SELECT ID, Name, Path, Date, Description FROM Pics"; if (!($connection = @ mysql_pconnect($hostName, $username, $password))) showerror(); if (!mysql_select_db("fermpics", $connection)) showerror(); if (!($result = @ mysql_query ($query, $connection))) showerror(); ?> <h1>Image database</h1> <h3>Click <a href="insert.php">here</a> to upload an image.</h3> <?php //require 'disclaimer'; if ($row = @ mysql_fetch_array($result)) { ?> <table> <col span="1" align="right"> <tr> <th>File Name</th> <th>Date</th> <th>Image</th> </tr> <?php do { ?> <tr> <td><?php echo "{$row["Name"]}";?></td> <td><?php echo "{$row["Date"]}";?></td> <td><?php echo "<img src=\"C:\wamp\www\fermpix\Pics\'{$row["Name"]}'\">";?></td> </tr> <?php } while ($row = @ mysql_fetch_array($result)); ?> </table> <?php } // if mysql_fetch_array() else echo "<h3>There are no images to display</h3>\n"; ?> </body> </html> I would appreciate any help. Cheers Paul I have uploaded a png image file to my server using FTP, the file is simply '.png' but displays the broken image icon
the image url is: http://www.instaworldcup.com/.png
how do i get this image to display with the name '.png' as I have seen on other sites, but i am having no luck
This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=332434.0 Hey guys - I have some code that pulls an image out of a folder and displays it on the page. The problem is, there won't always be an image to display in which case I'd rather the code not even display. Here's my code so far: Code: [Select] <tr><td><a href="images/uploads/<?php echo $row['loc_id']; ?>_1.jpg"><img id="morephoto" src="images/uploads/thumbs/<?php echo $row['loc_id']; ?>_1thb.jpg" /></a></td></tr> The images are renamed on upload to have the id number for that row appended to the front of the file name and that is how I'm calling them back in. I know I need to write an if statement that contains the code from the <tr> to the </tr> to display if the image exists, the problem is, since there isn't a field for this in the database, I don't know how to check it? I'm still a noob so I appreciate any help that is offered. Thanks! i am trying to display all the image from a dir and let the use lick on a image that takes them to other page where they can rename that image. or let them rename the image from the first page. here is what i have. Code: [Select] <?php SESSION_START(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <meta name="author" content="vs" /> <meta name="robots" content="index, follow" /> <title></title> </head> <body> <div id="centerColumn"> <div id="header"> </div> <p><span class="small"></span><br /> </p><br> <form method="post" action="data2file.php"> <?php echo ' <table> <tr><td>'; $username = $_SESSION['user']; $url = $username."./"; $handle = opendir ($url); while (false !== ($file = readdir($handle))) { if($file != "." && $file != ".." && $file != basename(__FILE__)) { //echo '<table border=1><tr><td>'; echo '<a href="edit.php" ><img border=1 src="'.$file.'"></a>'; echo '<br />'; echo "<input type=submit name=$file value=Edit>" ; print $file; echo '<br>'; echo "Enter new name: <input type='text' name='new-name' length='30'> <textarea type=text size='1' height='1'style='display:none' name='file-name' >". $file ."</textarea>"; echo '<hr>'; //echo '</td></tr></table>'; } } echo '</td>'; echo ' </table>'; ?> </body> </html> it works i loads all the images. and here is the data2file.php code Code: [Select] <?php SESSION_START(); echo $_POST['new-name']; echo '<br>'; echo $_POST['file-name']; echo '<br>'; echo "Successfully!"; ?> it's just display the name of the file and the new name i want to give it. But it only does the last image in the dir. How can i make the first code work so if the user clicks on the edit button it will pull only that image file name to the data2file.php and then from the data2file.php it will rename the file name. I am trying to display an image stream from my pond cam... The stream generated by my ip video server is a jpeg stream without boundaries. Causeur of this the normal mime replace isn't working. I need php to read a picture from the stream, and display the picture in the browser while reading the next picture in the background. I've got to the point that I am able to connect to the stream, read the first image, and display it in the browser as can be seen in the link below (cam is only on from 09:00 until 16:00) http://cam.xsiteit.nl/readstream_test.php From this point I need to read the second image and replace the image in the browser with the new one. This process has to continu until the browser is closed. Any help would be great! Regards, Peter I have the code below displaying images and image names. I want these to display in a table 2 rows high by the needed number of columns to show all the images in the directory. I have no idea what to do. What I am getting now is a single column with each image in its own row. <?php $path = "./uploaded/"; $dir_handle = @opendir($path) or die("Unable to open folder"); while (false != ($file = readdir($dir_handle))) { if($file == "index.php") continue; if($file == ".") continue; if($file == "..") continue; //show in a table 2 rows by required/needed number of columns echo'<div>'; echo '<table border="1">'; echo "<img src='$path/$file' alt='$file'>"."<img src='$file' alt='$file'>"; echo'</table>'; echo '<div>'; } ?> Hi i am trying to display image using 'CAT_BIO' => (isset($_GET['c'])) ? '<br />' '<img src="images/' . $row['image'] . ' " width="150" height="200" />' '', but i get the error Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in i think a comma or something missing can anyone spot where i am going wrong?? the other way i tried is <img src=\"images/{$row['image']}\" width=\"150\" height=\"200\" />" this one give no error bt doesnt display the anything it might be cz wrong path how do i make to go out one folder the get inside images folder?? not sure if its the solution help Well the subject line is pretty explicit. I found this script that uploads a picture onto a folder on the server called images, then inserts the the path of the image on the images folder onto a VACHAR field in a database table. Code: [Select] <?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{ // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "images/"; // Get our POSTed variable $image = $_FILES['image']; // Sanitize our input $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved to // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: member.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: member.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: member.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: images.php"); echo "File uploaded"; exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: member.php"); exit; } } //End of if session variable id is not present. ?> The script seems to work fine because I managed to upload a picture which was successfully inserted into my images folder and into the database. Now the problem is, I can't figure out exactly how to write the script that displays the image on an html page. I used the following script which didn't work. Code: [Select] //authenticate user //Start session session_start(); //Connect to database require ('config.php'); $sql = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' "); $row = mysql_fetch_assoc($sql); $imagebytes = $row['image']; header("Content-type: image/jpeg"); print $imagebytes; Seems to me like I need to alter some variables to match the variables used in the insert script, just can't figure out which. Can anyone help?? Hi, I am a newbie to this forum and php as well. I have a script which inserts data and image file into mysql database. Now I want to display this image on web page but unable to do it. I tried a lot but couldn't find out any solution. I would appreciate your help. <?php $con = mysql_connect("localhost", "root", "xx"); if(!$con) { die('Could not connect:' .mysql_error()); } mysql_select_db("yyy",$con); $memberid = '$_POST[memberid]'; $fname = '$_POST[fname]'; $lname = '$_POST[lname]'; $gender= '$_POST[gender]'; $add1 = '$_POST[add1]'; $add2 = '$_POST[add2]'; $city = '$_POST[city]'; $state = '$_POST[state]'; $zip = '$_POST[zip]'; $country = '$_POST[country]'; $photo1 = '$_FILES[image][name]'; echo "$photo1"; //if ($memberid= $_GET['memberid']){ $query="select memberid,fname,lname,gender,add1,add2,city,state,zip,country,photo1 from vadhuvar_member where memberid =(select max(memberid)from vadhuvar_member)"; $result=mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_assoc($result)) { $memberid = $row['memberid']; $fname = $row['fname']; $lname = $row['lname']; $gender = $row['gender']; $add1 = $row['add1']; $add2 = $row['add2']; $city = $row['city']; $state = $row['state']; $zip = $row['zip']; $country = $row['country']; //$photo1=chunk_split(base64_encode($photo1)); $photo1 = $row['photo1']; //echo base64_encode($photo1); echo "Name:$fname $lname<br>"; //echo "Name:$_POST[fname]$_POST[lname]<br>"; echo "Gender:$gender<br>"; echo "Address:$add1 $add2<br>"; echo "$city $state $zip<br>"; echo "$country<br>"; //echo "<img src=".upload/$photo1 ." alt="" width="200" height="300" /><br />"; } //} ?> <!--img src="<?php echo '.upload/$_FILES[image][name]?memberid=$row[memberid]'?>"alt="" width="200" height="300"/-->; <?php echo "<img src=". upload/$row[photo1]" alt="" width="200" height="300">";?> <!--img src="<?php echo ".\upload\$photo1" ?>" alt="" width="200" height="300"/-->; <? set_magic_quotes_runtime(1); // turn back on ?> Thanks Smita I have this software that lets user's use themes and so on. But one feature they didn't add into the peice of code is image display of the theme before the person chooses. I was hoping someone could help me figure out what bit of code I need to make it work. I know joomla has this feature but I don't care for joomla and I like this script alot more. All I can offer is the code where it displays the theme selector. PHP Part // HANDLE SUBMITTED DISPLAY SETTINGS // if (isset($_POST["SubmitDisplay"])) { myQ(" UPDATE `[x]users` SET `use_theme` = '{$_POST["theme"]}', WHERE `id`='".me('id')."' LIMIT 1 "); } /* Populate the site themes */ if ($handle = opendir("theme/")) { $themeInUse = (me("use_theme") != "" ? me("use_theme") : $CONF["DEFAULT_THEME"]); while (false !== ($file = readdir($handle))) { if ($file != "." and $file != ".." and $file != "templates" and is_dir("theme/".$file)) { $themesReplacementArray[] = array( "theme" => $file, "select" => ($file == $themeInUse ? "selected=\"selected\"" : ""), ); } } closedir($handle); } $tpl -> Loop("themes", $themesReplacementArray); HTML Part Code: [Select] <form method="post"> <table width="100%" border="0" cellspacing="2" cellpadding="0"> <tr> <td><strong>Site style: </strong></td> <td width="10"> </td> <td> <select name="theme" id="theme"> <loop themes> <option value="{theme}" {select}>{theme}</option> </loop themes> </select> </td> </tr> <tr> <td colspan="3"><input name="SubmitDisplay" type="submit" class="submit" id="SubmitDisplay" value="Save" /></td> </tr> </table> </form> Thanks in advance. |
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