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PHP - Uploaded Image, Change Width And Height
Hey guys!
I have the following php code that grabs variables (and the browsed image) from Flash. //FLASH VARIABLES $Name = $_POST['Name']; $itemNumber = $_POST['itemNumber']; $filename = $_FILES['Filedata']['name']; $filetmpname = $_FILES['Filedata']['tmp_name']; $fileType = $_FILES["Filedata"]["type"]; $fileSizeMB = ($_FILES["Filedata"]["size"] / 1024 / 1000); list($filename, $extension) = explode('.', basename($_FILES['Filedata']['name'])); $filename = $Name; $target = $filename . $itemNumber . "." . $extension; // Place file on server, into the images folder move_uploaded_file($_FILES['Filedata']['tmp_name'], "images/".$target); This works perfect, but what I want to change is the width and height of the uploaded image. Any ideas/suggestions on how this could be done? Thanks in advance!! Cheers! Similar Tutorialsi am uploading image to s3 server and by php i am making a copy of that image on server , if i upload an image of 2.3 mb than the width of image is not coming but if i upload less size image like 26kb than it is showing the width of image so it is able to create the copy . here is my code of php : $s3 = new S3(awsAccessKey, awsSecretKey); $thumbId = uniqid(); $thumbId .= ".jpg"; $img = ''; if($imgType == "image/jpeg"){ $img = imagecreatefromjpeg($sourceUrl); }else if($imgType == "image/png"){ $img = imagecreatefrompng($sourceUrl); }else if($imgType == "image/gif"){ $img = imagecreatefromgif($sourceUrl); }else{ $img = imagejpeg($sourceUrl); } echo $width = imagesx( $img ); echo $height = imagesy( $img ); please tell me what is the problem with size of image.. regards rahul Hi Fellas, having a bit of a brain freeze here, i am trying to set a max width or height for a uploaded image, but my brain has frozen, maybe its the cold weather lol list($width,$height)=getimagesize($uploadedfile); $newwidth=250; $newheight=($height/$width)*$newwidth; $tmp=imagecreatetruecolor($newwidth,$newheight); basicly i want to find the biggest size of an image, beit width or height then set that size to 250... Any pointers here? Hey guys I have a small issue, i have an upload that resizes the image into thumbnail by max width and ratios the width based on that. Here is my code
What I am wanting to do is instead upload the image with a max height and ratio the width proportionally. What variables do I have to reverse? Okay, so here is the deal. Have a table which stores image as blob files. Now i want to read the image width and height directly from the blob field. Is this possible and if yes, how? Things i tried so far; list($size[0],$size[1],$type, $attr) = getimagesize('image.php?i=26ddd45b02859e836d13d4b9fde34281'); print_r($size); $img = 'image.php?i=26ddd45b02859e836d13d4b9fde34281'; echo imagesy($img); image.php grabs the image from DB and show's it with header("Content-type: image/jpg"); It works for just showing the images with the <img> tag. Any ideas of help would be great! I have folder with 100's images, how to find Height and width as Excel sheet? Hi I need to change the name of a file being uploaded by a user. The reason i need this is because there is a strong possibility that duplicate filenames would be logged. This is the code i have currently: Code: [Select] $upload_path = 'cv/'; // The place the files will be uploaded to (currently a 'files' directory). $filename = $_FILES['userfile']['name']; // Get the name of the file (including file extension). $ext = substr($filename, strpos($filename,'.'), strlen($filename)-1); // Get the extension from the filename. // Check if we can upload to the specified path, if not DIE and inform the user. if(!is_writable($upload_path)) die('You cannot upload to the specified directory, please CHMOD it to 777.'); // Upload the file to your specified path. if(move_uploaded_file($_FILES['userfile']['tmp_name'],$upload_path . $filename)); This code works fine to upload the file in the current name. I assume i need to seperate the filename from the file extension, and i can then assign a new variable to the filename. Easier said than done though as Ive tried many combinations of things. Is there a simple way using this script? or will i need to start from scratch? Cheers Hi. I have a script here that will let users upload an image to my website but I just can't figure out how to save the uploaded image as "upload/logo.png" so that it will replace the already existing "upload/logo.png". Help would be greatly appreciated. Code: [Select] <html> <body> <form action="" method="post" enctype="multipart/form-data"> <label for="file">Filename:</label> <input type="file" name="file" id="file" /> <br /> <input type="submit" name="submit" value="Submit" /> </form> </body> </html> <?php if(isset($_POST['submit']) && !empty($_FILES["file"]["name"])) { $timestamp = time(); $target = "upload/"; $target = $target . basename($_FILES['uploaded']['name']) ; $ok=1; $allowed_types = array("image/gif","image/jpeg","image/pjpeg","image/png","image/bmp"); $allowed_extensions = array("gif","png","jpg","bmp"); if ($_FILES['file']['size'] > 350000) { $max_size = round(350000 / 1024); echo "Your file is too large. Maximum $max_size Kb is allowed. <br>"; $ok=0; } if ($_FILES["file"]["error"] > 0) { echo "Error: " . $_FILES["file"]["error"] . "<br />"; $ok=0; } else { $path_parts = pathinfo(strtolower($_FILES["file"]["name"])); if(in_array($_FILES["file"]["type"],$allowed_types) && in_array($path_parts["extension"],$allowed_extensions)){ $filename = $timestamp."-".$_FILES["file"]["name"]; echo "Name: " . $filename . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; $path_parts = pathinfo($_FILES["file"]["name"]); echo "Extension: " . $path_parts["extension"] . "<br />"; echo "Size: " . round($_FILES["file"]["size"] / 1024) . " Kb<br />"; //echo "Stored in: " . $_FILES["file"]["tmp_name"]. " <br />"; } else { echo "Type " . $_FILES["file"]["type"] . " with extension " . $path_parts["extension"] . " not allowed <br />"; $ok=0; } } if($ok == 1){ @move_uploaded_file($_FILES["file"]["tmp_name"], $target . $filename); $file_location = $target . $filename; if(file_exists($file_location)){ echo "Uploaded to <a href='$file_location'>$filename</a> <br />"; } else { echo "There was a problem saving the file. <br />"; } } } else { echo "Select your file to upload."; } ?> Thanks! Hi there. How can i set up a maximum width and height of a picture? For example: the limit is 400x400. So if the photo is 200x200 then it stays that way, however if a photo is 550x550 then it gets resized to 400x400. Hello I have uploaded images from a server onto a website, they run through a mysql database (database holds the image file names). The problem I am having is resizing the images. Can anyone help? Thank you in advance GWG Hello. I'm using an Amazon S3 class to uploaded to S3. I have 2 upload boxes - The first uploads once and the second needs to upload twice - 1 full size, 1 thumb. The issue i'm having is that the 2nd image (the thumb) seems to be failing, although if I don't save the first full sized image I am able to upload the thumb. So I think the issue is with using the temp file twice? This is my code: //retreive post variables $fileName = $randomString . "_" . $_FILES['theFile']['name']; $fileTempName = $_FILES['theFile']['tmp_name']; $fileName2 = $randomString . "_" . $_FILES['theFile2']['name']; $fileTempName2 = $_FILES['theFile2']['tmp_name']; //move the file if ($s3->putObjectFile($fileTempName, "containerhere", $fileName, S3::ACL_PUBLIC_READ)) { echo "<strong>Uploaded Image</strong>"; }else{ echo "<strong>Something went wrong while uploading your file... sorry.</strong>"; } //move the file if ($s3->putObjectFile($fileTempName2, "containerhere", $fileName2, S3::ACL_PUBLIC_READ)) { echo "<strong>Uploaded Image</strong>"; }else{ echo "<strong>Something went wrong while uploading your file... sorry.</strong>"; } include('simpleImage.php'); $image = new SimpleImage(); $image->load($_FILES['theFile2']['tmp_name']); $image->resizeToWidth(100); $image->save($_FILES['theFile2']['tmp_name']); $fileName3 = $randomString . "_" . $_FILES['theFile2']['name']; $fileTempName3 = $_FILES['theFile2']['tmp_name']; //move the file if ($s3->putObjectFile($fileTempName3, "containerhere", "thumbs/" . $fileName3, S3::ACL_PUBLIC_READ)) { echo "<strong>Uploaded Image</strong>"; }else{ echo "<strong>Something went wrong while uploading your file... sorry.</strong>"; } Can anyone offer any advice? Thanks. Hi, I have a script that uploads an image to a directory and then saves the fill path to a field in a table for later use. The only problem is people are uploading huge images and then when I produce a catalogue of my listings it takes forever to load because the images are so big. What I am after is an add in script to create an ADDITIONAL image 100 x 75px, I don't really want to change my upload script. Any ideas? Thanks in advace. Here is what I have: Code: [Select] <?php $idir = "../fleet/"; // Path To Images Directory if (isset ($_FILES['fupload'])){ //upload the image to tmp directory $url = $_FILES['fupload']['name']; // Set $url To Equal The Filename For Later Use if ($_FILES['fupload']['type'] == "image/jpg" || $_FILES['fupload']['type'] == "image/jpeg" || $_FILES['fupload']['type'] == "image/pjpeg") { $file_ext = strrchr($_FILES['fupload']['name'], '$account.'); // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php $copy = copy($_FILES['fupload']['tmp_name'], "$idir" . $_FILES['fupload']['name']); // Move Image From Temporary Location To Perm } } $fleetimage1 = mysql_real_escape_string("$idir" . $_FILES['fupload']['name']); //then insert sql code below... ?> I'm beginning with simple codes to understand how things work and work my way up with what I need. My upload form looks like this: <form name="upload about" action="display.php" enctype="multipart/form-data" method="post"> Header: <input type="text" name="header" /> Body: <input type="text" name="body" /> <input type="hidden" name="MAX_FILE_SIZE" value="30000" /> Image: <input type="file" name="pic" /> <input type="submit" /> </form> And my display page is just as simple. <?php echo $_POST["header"]; ?> <?php echo $_POST["body"]; ?> <?php echo $_FILES['pic']['name']; ?> The problem is, whenever I submit the data, everything is fine, except the image does not display and just posts the filename of the image I uploaded. According to my research setting the enctype to multipart/formdata should display the image, but it does not. Can anyone tell me what's wrong? Hi everyone, I have a script below, which uplads an image, however, the image name always starts with a capital letter, I want all letters to be small, how to adjust this please, thank you
$target_dir = "";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$target_file = "$get_current_user.jpg";
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Check if image file is a actual image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo ""; /// was File is an image
$uploadOk = 1;
} else {
echo "File is not an image<br>";
$uploadOk = 0;
}
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 10000000) {
echo "Sorry, this image is too large, please resize using Paint<br>";
$uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg" && $imageFileType != "gif" ) {
echo "Only JPG, JPEG, PNG & GIF files are allowed<br>";
$uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
echo "There was an error<br>";
} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "Uploaded successfully (You may need to clear Cache to see the new image)";
mysql_query("UPDATE user SET user_image = 'https://.../images/users/$get_current_user.jpg' WHERE user_name = '$get_current_user' ");
} else {
echo ""; /// was Select a suitable image, file not uploaded yet
}
}
Not sure if this is the right place to post this. I have PHP form that I use to upload a document, PDF or Word Doc, I would also like the form to create a thumbnail of the document when it is uploaded, is this possible? Hi Basically I've built a CMS where by my clients can upload a number of images. On the success page I want to display the images they uploaded by file name. The issue is the number of images can vary. They may upload 2 or 10 or 50 etc. So far I've come up with this: Code: [Select] // number of files $UN = 3; //I've set this to 3 for now, but this is passed from the upload page! // server directories and directory names $dir = '../properties'; $images = glob($dir.'/*.{jpg}', GLOB_BRACE); //formats to look for $num_of_files = $UN; //number of images to display from number of uploaded files foreach($images as $image) { $num_of_files--; $newest_mtime = 0; $image = 'BROKEN'; if ($handle = @opendir($dir)) { while (false !== ($file = readdir($handle))) { if (($file != '.') && ($file != '..')) { $mtime = filemtime("$dir/$file"); if ($mtime > $newest_mtime) { $newest_mtime = $mtime; $image = "$file"; } } } } if($num_of_files > -1) //this made me laugh when I wrote it echo $trimmed = ltrim($image, "../properties").'<br />'; //display images else break; } Without this piece of code: Code: [Select] $newest_mtime = 0; $image = 'BROKEN'; if ($handle = @opendir($dir)) { while (false !== ($file = readdir($handle))) { if (($file != '.') && ($file != '..')) { $mtime = filemtime("$dir/$file"); if ($mtime > $newest_mtime) { $newest_mtime = $mtime; $image = "$file"; } } } } It shows the first 3 files alphabetically. I want to view the last number of images added. With the above code it simply shows the last image added 3 times! So I need to get the time each image was added and then order by the newest added and limit to the number of images uploaded. Any suggestions please? Kindest regards Glynn hi to all.Im currently displaying the images from my upload directory but the image does not display.please help thanks OK, When the user fills the info in the form out it goes in the DB fine. I can then array them on the "showroom page" fine. When they upload a picture it goes into the /images/ folder fine. Problem is... On each array on the showroom page I need the image they uploaded to be displayed. Cant work it out. Help would be GREAT!!!!! Hi People. Thanks for all the help on here in the past, you have been brilliant. I run the airfield cards.com website and have a script in the output page that is as follows. Code: [Select] To Embed this Card into your website, cut and paste the following code: <input name="generate" value="<? echo "<iframe src='http://www.airfieldcards.com/php/courtesy_card.php?id=".$id. "' ></iframe>";?>"> Now the way it shows on the site is like this Code: [Select] <iframe src='http://www.airfieldcards.com/php/courtesy_card.php?id=165' ></iframe> So, I look at the iframe code from google that should be cut and pasted into a website (obviously it's working code from google) Code: [Select] <iframe width="700" height="1500" frameborder="0" scrolling="no" marginheight="0" marginwidth="0" src="http://maps.google.co.uk/maps/ms?msa=0&msid=203054009368331462137.0004aa731e7de2e86a0e0&ie=UTF8&t=h&vpsrc=0&ll=53.800651,-4.042969&spn=80.03397,61.435547&z=4&output=embed"></iframe><br /><small>View <a href="http://maps.google.co.uk/maps/ms?msa=0&msid=203054009368331462137.0004aa731e7de2e86a0e0&ie=UTF8&t=h&vpsrc=0&ll=53.800651,-4.042969&spn=80.03397,61.435547&z=4&source=embed" style="color:#0000FF;text-align:left">AirfieldCards.com</a> in a larger map</small> Now I have tried to take the code from the google iframe (width/height/scrolling/etc) and paste it into my source but it doesn't work. Here's how I have done it. Code: [Select] To Embed this Card into your website, cut and paste the following code: <input name="generate" value="<? echo "<iframe width="700" height="1500" frameborder="0" scrolling="no" marginheight="0" marginwidth="0" src='http://www.airfieldcards.com/php/courtesy_card.php?id=".$id. "' ></iframe>";?>"> Can someone please modify my fist lot of code so that I have the correct height and width settings (700 x 1500) Thanks in advance Regards Vince Gledhill I need help making the uploaded image file name, that's chosen to be uploaded, be displayed on the html page with the path /upload/ added to the beginning of the displayed file name like so: ../upload/test.png Any help/improvements will be appreciated.
<html>
<head>
<title>PHP Test</title>
</head>
<body>
<?php
if ($form_submitted == 'yes') {
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = strtolower( end($temp) );
if ( $_FILES["file"]["size"] < 200000
&& in_array($extension, $allowedExts) )
{
if ($_FILES["file"]["error"]!= 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
} else {
$length = 20;
move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $newfilename );
$file_location = '<a href="http://../upload/' . $newfilename . '">' . $newfilename . '</a>';
}
} else {
echo "Invalid upload file";
}
?>
<label for="file">Filename:</label>
<input type="file" name="file" id="file">
</body>
</html>
Hi folks, I'm using this code to display a picture that is uploaded in a form. Code: [Select] print "<img src='http://www.coast2kosci.com/mylongrun/test/photo_uploads/$imgx' width='300' height='300' alt='Your photo' />"; At first I tried specifying only the width, without the height. But for tall photos, the display wasn't really what I was after. Then I specified both width and height, which results in a mis-scaled photo. Is there any way that I can specify the height and width atributes as a sort of maximum / or limiter. Ie: If the height is greater than the width, limit the height to 300, and if the width is greater than the height, limit the width to 300? Thanks for your time, Dave |
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