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PHP - Gd Images
Ok, I have a GD image that is then rewrote as a png. My question is, is it possible to find out what site is requesting the image and then use it on the php image. Example, if I didn't want a domain to use my images, could I somehow me able to restirict that site?
Similar TutorialsI have this script from http://lampload.com/...,view.download/ (I am not using a database) I can upload images fine, I can view files, but I want to delete them. When I press the delete button, nothing happens
http://www.jayg.co.u...oad_gallery.php
<form>
<?php $dir = dirname(__FILENAME__)."/images/gallery" ; $files1 = scandir($dir); foreach($files1 as $file){ if(strlen($file) >=3){ $foil = strstr($file, 'jpg'); // As of PHP 5.3.0 $foil = $file; $pos = strpos($file, 'css'); if ($foil==true){ echo '<input type="checkbox" name="filenames[]" value="'.$foil.'" />'; echo "<img width='130' height='38' src='images/gallery/$file' /><br/>"; // for live host //echo "<img width='130' height='38' src='/ABOOK/SORTING/gallery-dynamic/images/gallery/ $file' /><br/>"; } } }?> <input type="submit" name="mysubmit2" value="Delete"> </form>
any ideas please?
thanks
Hey every i'm just trying to change my 'Read More' text after a blog post on my site into am image. Got some code that lets me change the actual text to something else, just can't find how to change the text into an image. Code: [Select] add_filter( 'the_content_more_link', 'child_content_more_link', 10, 2 ); function child_content_more_link( $more_link, $more_link_text ) { return str_replace( $more_link_text, 'my new read more text', $more_link ); } Thanks, Matty44m Hi guys I just want to know something. With php you can handle excel files by saving them as .csv files. Then the php script reads it line by line, and commas separating the fields. But is it possible that I can insert an image (small jpg or png) file in one of these fields...and then handle the image in the php script? Just for intrest I inserted an image in excel worksheet and it did not insert it into a cell, it float. Is there maybe another to insert it in the cell and then control it with php?? Thanx in advance Hi, I have a pretty simple application that allows my users to upload images and i'm able to succesfull upload images to a new dir to be outputted. But as to save on download time and server space is there a simple piece of code I can add to the below code that will resize every image on uploded to 460(w) x 300 (h) from my users Code: [Select] <?php if ((($_FILES["article_image"]["type"] == "image/gif") || ($_FILES["article_image"]["type"] == "image/jpeg") || ($_FILES["article_image"]["type"] == "image/pjpeg")) && ($_FILES["article_image"]["size"] < 5000000)) { move_uploaded_file($_FILES["article_image"]["tmp_name"], "images/" . $_FILES["article_image"]["name"]); } ?> Thanks in advance for any help and suggestions please help me to solve this problem : when my client want to remove a record all the related images must be deleted with it ,is that possible?? Ok, so i've managed to upload images, but now i'm trying to rename them to avoid replacement and i've done something like this: mysql_query( "INSERT INTO `softmarket`.`internet_securitate` (`internet_securitateID`, `nume_produs`, `descriere`, `versiune`,`poza`, `pret`, `disponibilitate_produs`) VALUES ('', '$_POST[titlu]', '$_POST[descriere]', '$_POST[versiune]', '$foto', '$_POST[pret]', '$_POST[disponibilitate]');"); Code: [Select] function findexts ($filename) { $filename = strtolower($filename) ; $exts = split("[/\\.]", $filename) ; $n = count($exts)-1; $exts = $exts[$n]; return $exts; } $ext = findexts ($_FILES['foto']['name']) ; $ran2 = $foto."."; $target = "../images/"; $target = $target . $ran2.$ext; while (file_exists($target)) { $ran2 = 'copyof'.$ran2; $target = $target . $ran2.$ext; } if(move_uploaded_file($_FILES['foto']['tmp_name'], $target)) { echo "The file has been uploaded as ".$ran2.$ext; } else { echo "Sorry, there was a problem uploading your file."; } header('Location: succes.php'); but it renames them in a wierd way...what is wrong? Hi all, while I was coding a script for my website something struck me which ive not really got an idea how to code it, but anyway, in my Database ive got a table which holds image URL's but how could I echo them out in PHP which will show the image? Not just the writing. Thanks for your help. Hi, There are 10 images in a folder called "Images". All images are named like 1.jpg, 2.jpg, 3.gif, 4.png, and so on... I am using these images on my website with this link: MyDomain.com/Images/1.jpg Now what i want to do is change the image every minute. from 1... to ...10 all images one by one, every minute... Link should remain same, but it should give different image every time. and if someone copies my link of image, it still gets changed every minute. i hope to get some help, better with some example code... Thanks! Hello all , I am facing a problem when trying to to show image files generated from php code in IE8 . it is working fine on firefox and IE9 . but in IE8 its showing the famous red X instead of the image ; belw is the code: Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html dir="RTL" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:v="urn:schemas-microsoft-com:vml" lang="en-us" > <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/> <head> <title>header </title> <link rel="stylesheet" type="text/css" href="style.css" /> </head> <body> <?php mb_internal_encoding('UTF-8'); /** * @author IT Department * @copyright 2011 */ INCLUDE ("header.php"); INCLUDE ("connect.php"); if (isset($_GET['id'])) { $ID=$_GET['id']; $sql="select employees.USER_ID, employees.NAME_ARAB, employees.NAME_ENG, employees.TITLE_ARAB, employees.TITLE_ENG, department.DEP_NAME_ENG, department.DEP_NAME_ARAB, location.LOC_DESC_ENG, location.LOC_DESC_ARAB, employees.MOBILE, employees.EMAIL, employees.DEP, employees.LOC FROM employees,location,department WHERE employees.DEP=department.DEP_ID AND employees.LOC=location.LOC_ID and employees.USER_ID=$ID "; //echo $sql; $result=mysql_query($sql,$link); mysql_query("SET NAMES 'utf8'"); ECHO '<html DIR="RTL">'; echo "<table width=100% border='1' cellpadding='8' align='center' bgcolor='white'>"; while ($row = mysql_fetch_array($result, MYSQL_NUM)) { //$row = mysql_fetch_assoc($result); echo"<tr align='center'>"; echo"<td>"; ECHO " الرقم الوظيفي :"; echo"</td>"; echo"<td colspan='2'>"; ECHO "<strong>$row[0]</strong>"; echo"</td>"; echo"<td>"; ECHO " : User ID "; echo"</td>"; echo"<td rowspan='7'>"; ECHO "<img src='IMAGE.php?id=".$row[0]."' width='225' height='300'/>"; //the image echo"</td>"; echo "</tr>"; the php image file is : Code: [Select] <?php INCLUDE ("connect.php"); $ID = "'".$_GET['id']."'"; $sql="SELECT PHOTO FROM employees WHERE USER_ID=$ID"; $result=mysql_query($sql,$link); $row = mysql_fetch_array($result); header("Content-type: image/pjpeg"); print ( $row[0]); ?> Thanks in advance for your help I have a site where I need to have lets call it image1 displayed, then I want to change this image based on a php if statement, for instance: if $var == $var2 change the image ....blah blah so I was also going to have the names of my images stored in my database, i.e. image1.jpg and image2.jpg in my database. the image is in its own div tag set as the background image of the div tag if that makes any diference. Thanks Ok I have finally got rid of my syntax errors with this upload script. Now im getting the error uploading your image. My folders are called project_files and project_thumbs. This is my script <?php $results = ""; error_reporting(E_ALL); ini_set("display_errors", 1); //Auto Thumbnail Function function create_thumbnail($source,$destination, $thumb_width) { $size = getimagesize($source); $width = $size[0]; $height = $size[1]; $x = 0; $y = 0; if($width> $height) { $x = ceil(($width - $height) / 2 ); $width = $height; } elseif($height> $width) { $y = ceil(($height - $width) / 2); $height = $width; } $new_image = imagecreatetruecolor($thumb_width,$thumb_width) or die('Cannot initialize new GD image stream'); $extension = get_image_extension($source); if($extension=='jpg' || $extension=='jpeg') $image = imagecreatefromjpeg($source); if($extension=='gif') $image = imagecreatefromgif($source); if($extension=='png') $image = imagecreatefrompng($source); imagecopyresampled($new_image,$image,0,0,$x,$y,$thumb_width, $thumb_width,$width,$height); if($extension=='jpg' || $extension=='jpeg') imagejpeg($new_image,$destination); if($extension=='gif') imagegif($new_image,$destination); if($extension=='png') imagepng($new_image,$destination); } //Get Extension Function function get_image_extension($name) { $name = strtolower($name); $i = strrpos($name,"."); if (!$i) { return ""; } $l = strlen($name) - $i; $extension = substr($name,$i+1,$l); return $extension; } //Random Name Function function random_name($length) { $characters = "abcdefghijklmnopqrstuvwxyz01234567890"; $name = ""; for ($i = 0; $i < $length; $i++) { $name .= $characters[mt_rand(0, strlen($characters) - 1)]; } return "image-".$name; } $images_location = "/project_files/"; $thumbs_location = "/project_thumbs/"; $thumb_width = "100"; $maximum_size = "5000000"; //Check And Save Image if($_POST) { if($_FILES['image']['name'] == ""){ $results = "Please select the image you would like to upload by clicking browse"; } else{ $size=filesize($_FILES['image']['tmp_name']); $filename = stripslashes($_FILES['image']['name']); $extension = get_image_extension($filename); if($size > $maximum_size) { $results = "Your file size exceeds the maximum file size!"; } else if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { $results = "Only the following extensions are allowed. 'jpg', 'jpeg', 'png', 'gif'."; } else { $image_random_name=random_name(15).".".$extension; $copy = @copy($_FILES['image']['tmp_name'], $images_location.$image_random_name); if (!$copy) { $results = "Error while uploadin your image! Please try again!"; } else{ create_thumbnail($images_location.$image_random_name,$thumbs_location.$image_random_name, $thumb_width); $results = "Your image has been uploaded!"; } } } } ?> and the form Code: [Select] <html> <head> <title>test</title> </head> <body> <?php echo $results; ?> <form action="#" method="post" enctype="multipart/form-data"> <input type="hidden" name="MAX_FILE_SIZE" value="5000000" /> <input type="file" name="image" /> <input type="submit" value="Upload Image" /> </form> </body> </html> This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=346417.0 Hi, I have this upload script that uploads a file and writes path to DB: Code: [Select] <?php $uploadDir = '../uploads/'; if(isset($_POST['upload'])) { foreach ($_FILES as $file) { $fileName = $file['name']; $tmpName = $file['tmp_name']; $fileSize = $file['size']; $fileType = $file['type']; if($fileName==""){ $filePath = '../img/none.jpg'; } else{ $filePath = $uploadDir . $fileName; } if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); $filePath = addslashes($filePath); } $fileinsert[]=$filePath; } ?> this works fine for a single image, but I eed to upload 2 images with a single submission....any ideas how to do this? Thanks Hello, I have been wrestling with this forth past couple of days and cannot work out what is wrong. The code below works ok except when I wish to change the picture. Instead of changing the image with the corresponding images it uses the "nopic" image even though the correct image is still in the database. So for example if I enter some data with one image attached this appears correctly. If I then change the image the old image reverts to "nopic" and the new image is displayed with the latest data. Even though in the database the old image is still present Hope this makes sense What am I doing wrong? offending code below Any help greatly appreciated Code: [Select] // Connect to the database $dbc = mysqli_connect(DB_Host, DB_User, DB_Password, DB_Name); // Retrieve the user data from MySQL $query = "SELECT * FROM conversation where topic = 'polls' "; //$data = mysqli_query($dbc, $query); echo '<table>'; $result = mysqli_query($dbc, $query); while ($row = mysqli_fetch_array($result)) { if (is_file(MM_UPLOADPATH . $row['picture']) && filesize(MM_UPLOADPATH . $row['picture']) > 0) { echo '<tr><td class="label"></td><td><img src="' . MM_UPLOADPATH . $row['picture'] . '" alt="Profile Picture" style="width:75px; max-height:55px;" /> ' . $row["quote"] . ''; } else { echo '<tr><td class="label"></td><td><img src=" ' . MM_UPLOADPATH . 'nopic.png" style="width:60px; maxheight=44px;" /> ' . $row["quote"] . ''; } } echo '</tr>'; echo '</table>' ?> Hi guys, i have a form that uploads a school photo. All i want to do is resize the image to its widest dimension of a width of 480 px and then obviously constrain the height. can anyone help me, ive tried to do this myself but when it comes to image properties i really struggle etc. heres my upload code. Code: [Select] <?php include("includes/connection.php"); // Where the file is going to be placed $schoolimage = "SchoolImages/"; //This path will be stored in the database as it does not contain the filename $currentdir = getcwd(); $path = $currentdir . '/' . $schoolimage; // Get the schoolid for the image and school linker table $schoolid = $_POST['schoolid']; //Get the school name $query = "SELECT * FROM school WHERE school_id = ".$schoolid; $result = mysql_query($query) or die("Error getting school details"); $row = mysql_fetch_assoc($result); $schoolname = $row['name']; //Use this path to store the path of the file in the database. $filepath = $schoolimage . $schoolname; //Create the folder if it does not already exist if(!file_exists('SchoolImages')) { if(mkdir('SchoolImages')) { echo 'Folder ' . 'SchoolImages' . ' created.'; } else { echo 'Error creating folder ' . 'SchoolImages'; } } //Store the folder for the course title. if(!file_exists( $filepath )) { if(mkdir( $filepath )) { echo 'Folder ' . $schoolname . ' created.'; } else { echo 'Error creating folder ' . $schoolname; } } // Where the file is going to be placed $target_path = $filepath; // Add the original filename to our target path. Result is "uploads/filename.extension" echo $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"; $filename = $_FILES['uploadedfile']['name']; //Store the filename, path other criteria in the database echo $query = "INSERT INTO image(image_id, name, path) VALUES(0, '$filename', '$filepath')"; //Perform the query $add = mysql_query($query, $conn) or die("Unable to add the image details to the database"); $imageid = mysql_insert_id(); //Store the filename, path other criteria in the database echo $query = "INSERT INTO image_school( image_id, school_id ) VALUES('$imageid', '$schoolid')"; //Perform the query $add = mysql_query($query, $conn) or die("Unable to add the image details to the database"); $message = 'Upload Successful'; } else { $message = 'There was an error uploading the file, please try again!'; } //Close the connection to the database mysql_close($conn); header("Location: add_school_photo_form.php? message={$message}&schoolid={$schoolid}"); //header("Location: add_school_photo_form.php? message=$message, schoolid=$schoolid"); exit(); ?> Id be eternially gratefull. Kind Regards Dean I am looking to display image paths in a row separated by commas. There are 6 images that goes to each user and I would like only the 6 images at be in each " " Like this: "images/listings/listing_516013019A-only.jpg,images/listings/listing_848813453A-1.jpg,images/listings/listing_664613453A-2.jpg,images/listings/listing_520313453A-3.jpg,images/listings/listing_690513453A-4.jpg,images/listings/listing_125113453A-5.jpg,images/listings/listing_641013453A-6.jpg," "images/listings/listing_736913186A-1.jpg,images/listings/listing_822713186A-2.jpg,images/listings/listing_136513186A-3.jpg,images/listings/listing_700313186A-4.jpg,images/listings/listing_716013186A-5.jpg,images/listings/listing_213113186A-6.jpg," "images/listings/listing_292113254A..-1.jpg,images/listings/listing_854413254A..-2.jpg,images/listings/listing_446013254A..-3.jpg,images/listings/listing_676313254A..-4.jpg,images/listings/listing_563413254A..-5.jpg,images/listings/listing_341513254A..-6.jpg," Right now it is displaying them like this "images/listings/listing_516013019A-only.jpg," "images/listings/listing_848813453A-1.jpg," "images/listings/listing_664613453A-2.jpg," "images/listings/listing_520313453A-3.jpg," "images/listings/listing_690513453A-4.jpg," "images/listings/listing_125113453A-5.jpg," "images/listings/listing_641013453A-6.jpg," "images/listings/listing_736913186A-1.jpg," "images/listings/listing_822713186A-2.jpg," "images/listings/listing_136513186A-3.jpg," "images/listings/listing_700313186A-4.jpg," "images/listings/listing_716013186A-5.jpg," "images/listings/listing_213113186A-6.jpg," "images/listings/listing_292113254A..-1.jpg," "images/listings/listing_854413254A..-2.jpg," "images/listings/listing_446013254A..-3.jpg," "images/listings/listing_676313254A..-4.jpg," "images/listings/listing_563413254A..-5.jpg," "images/listings/listing_341513254A..-6.jpg," Here is the code I have: Code: [Select] <?php // Make a MySQL Connection mysql_connect("localhost", "xxxxxxxx", "xxxxxxxx") or die(mysql_error()); mysql_select_db("xxxxxxxx") or die(mysql_error()); $result = mysql_query("SELECT * FROM listimages ORDER BY listimages.listingid DESC ") or die(mysql_error()); while($row = mysql_fetch_array($result)) { echo "\""; echo "$row[imagepath],"; echo "\""; echo "<br>"; } ?> My tables for the images is "listimages" and the columns a id (which are the auto_increments) imagepath (which shows the path/image1.jpg) mainimage (which just shows 0 or 1 depending on the picture that is the default for that listing, 1 being default) listingid (shows numbers 1 2 3 etc corresponding to the Id in the listings table to show what images go with what listing) There are up to 6 images for each listing. Any idea how to fix this? Hi guys I have got this code here, giving quite a few errors and was wondering if anyone could help me sort it out, the script (should) run straight away just have to edit the database connect, I've tried manually creating the table and taking out the create table but when I try and upload an image I get these errors: Warning: move_uploaded_file(latest.img) [function.move-uploaded-file]: failed to open stream: Permission denied in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 23 Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move '/Applications/XAMPP/xamppfiles/temp/php2lgspw' to 'latest.img' in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 23 Warning: fopen(latest.img) [function.fopen]: failed to open stream: No such file or directory in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 24 Warning: filesize() [function.filesize]: stat failed for latest.img in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 25 Warning: fread() expects parameter 1 to be resource, boolean given in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 25 here is the code: Code: [Select] <?php // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); // First run ONLY - need to create table by uncommenting this // Or with silent @ we can let it fail every subsequent time ;-) $q = < < <CREATE create table pix ( pid int primary key not null auto_increment, title text, imgdata longblob) CREATE; @mysql_query($q); // Insert any new image into database if ($_REQUEST[completed] == 1) { // Need to add - check for large upload. Otherwise the code // will just duplicate old file ;-) // ALSO - note that latest.img must be public write and in a // live appliaction should be in another (safe!) directory. move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($image) < 149000) { mysql_query ("insert into pix (title, imgdata) values (\"". $_REQUEST[whatsit]. "\", \"". $image. "\")"); } else { $errmsg = "Too large!"; } } // Find out about latest image $gotten = @mysql_query("select * from pix order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } // If this is the image request, send out the image if ($_REQUEST[gim] == 1) { header("Content-type: image/jpeg"); print $bytes; exit (); } ?> <html><head> <title>Upload an image to a database</title> <body bgcolor=white><h2>Here's the latest picture</h2> <font color=red><?= $errmsg ?></font> <center><img src=?gim=1 width=144><br> <b><?= $title ?></center> <hr> <h2>Please upload a new picture and title</h2> <form enctype=multipart/form-data method=post> <input type=hidden name=MAX_FILE_SIZE value=150000> <input type=hidden name=completed value=1> Please choose an image to upload: <input type=file name=imagefile><br> Please enter the title of that pictu <input name=whatsit><br> then: <input type=submit></form><br> <hr> By Graham Ellis - graham@wellho.net </body> </html> I am designing a web site which will dynamically generate labels which will be printed out by the end user. I have coded the part which generates each label into an image, but sometimes a user will want to print multiple copies of the same label on a page. I am working on the basis of an A4 page initially and depending on the design of the label, they will be different sizes (but they wont be mixing different designs on a single page so for each a4 sheet, every label will be the same size). I have no idea how to go about the dynamic arranging of images onto the page without them being half chopped off. Ideally they will just be arranged onto a HTML page which I can just add a print button to. Any suggestions are most welcome Hello all I have a requirement to create images mostly made up of rectangles circles and squares etc I have discovered imageline but is there anything a bit more exspansive? I am absolutley stumped with this: im trying to have PHP get the size of an image that has been uploaded to mySQL, and return it to a certain size, lets say 200X200px. The problems im having to finding an existing script to help me understand - and then fit it into a table? All the help from google searches seem to suggest using software, but i would like to learn how to do this properly. First off the upload script: Code: [Select] <?php error_reporting(E_ALL); ini_set("display_errors", 1); echo '<pre>' . print_r($_FILES, true) . '</pre>'; //This is the directory where images will be saved $target = "/home/users/web/b109/ipg.removalspacecom/images/COMPANIES"; $target = $target . basename( $_FILES['upload']['name']); //This gets all the other information from the form $company_name=$_POST['company_name']; $basicpackage_description=$_POST['basicpackage_description']; $location=$_POST['location']; $postcode=$_POST['postcode']; $upload=($_FILES['upload']['name']); // Connects to your Database mysql_connect("server", "user", "password") or die(mysql_error()) ; mysql_select_db("removalspacelogin") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO `Companies` (company_name, basicpackage_description, location, postcode, upload) VALUES ('$company_name', '$basicpackage_description', '$location', '$postcode', '$upload')") ; echo mysql_error(); //Writes the photo to the server if(move_uploaded_file($_FILES['upload']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['upload']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } ?> this works great, it uploads the users pictures. This displays them out: Code: [Select] <?php $database="removalspacelogin"; mysql_connect ("server", "user", "password"); @mysql_select_db($database) or die( "Unable to select database"); $result = mysql_query( "SELECT company_name, location, postcode, basicpackage_description, premiumuser_description, upload FROM Companies" ) or die("SELECT Error: ".mysql_error()); $num_rows = mysql_num_rows($result); print "\n\n\nThere are $num_rows records.<P>"; echo "<table><tr><th>Comppany Name</th><th>Location</th><th>Postcode</th><th>Basic Members</th><th>Upgraded Users</th><th>Company Logo</th></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td></tr>";// store the records into $row array and loop through while ( $row = mysql_fetch_array( $result, MYSQL_ASSOC ) ) { // Print out the contents of the entry echo "<tr><td>{$row['company_name']}</td>"; echo "<td>{$row['location']}</td>"; echo "<td>{$row['postcode']}</td>"; echo "<td>{$row['basicpackage_description']}</td>"; echo "<td>{$row['premiumuser_description']}</td>"; echo "<td><img src=\"http://www.removalspace.com/images/COMPANIES{$row['upload']}\" alt=\"logo\" /></td></tr>";} echo "</table>"; ?> Right now all the information in my table is fine except the image coloumn, the images are all the same size as they went in, i'd like them to all be one size so there in a uniformed fashion. But need all the rest of the mySQL table displayed, so how does it fit into my script to display the table? Many thanks!??? |
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