I wanted to have a script which would turn all my images into thumbnails and save these new thumbnails in a new folder.
I had luck an found a code which worked almost perfectly from
http://webcheatsheet.com/php/create_thumbnail_images.php
The only problem is that if there is a folder in the "uploads"-folder (defined at the end of the code) then I get the "Notice: Undefined index: extension".
The code doesn't get stuck and I still get my thumbnails, but the error message is annoying.
I tried to put in an isset-function, but did something wrong as I still didn't manage to stop the script from acting upon the folders. The code doesn't react similarly on any other files, so it seems to be the lack of an extension in the folder-name that bothers the code.
I was able to make it easy and simply remove any folders from the "uploads"-folder and put the thumbnails' path somewhere else, but I'd also like to get it to work without error-messages in case I happen to have folders in these image folders.
// parse path for the extension
$info = pathinfo($pathToImages . $fname);
// continue only if this is a JPEG image
//print_r($info);
if ( strtolower($info['extension']) == 'jpg' ) { // reacts on the folder with no extension name and gives an error
echo "Creating thumbnail for {$fname} <br />";
// load image and get image size
$img = imagecreatefromjpeg( "{$pathToImages}{$fname}" );
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresized( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height );
// save thumbnail into a file
imagejpeg( $tmp_img, "{$pathToThumbs}{$fname}" );
}
}
The full code in the link above.
The pathinfo docs explain:
Note:
If the path does not have an extension, no extension element will be
returned
So to avoid the notice you just need to check that value is available before trying to use it:
if( isset($info['extension']) AND strtolower($info['extension']) == 'jpg'){
//do sutff
}
Or instead of isset(...) you could use array_keys_exists('extension', $info).
Related
When this code runs the browser screen turns black with a grey and white checkered square in the middle when it is suppose to display a thumbnail within a table cell on a form.
function createThumb( $imageUrl, $thumbWidth )
{
// load image and get image size
$img = imagecreatefromjpeg( $imageUrl );
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresized( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height );
// display thumbnail
header("Content-type:image/jpeg");
imagejpeg( $tmp_img,$imageUrl,80);
};
And the line that calls the code is
<td >
<?php echo createThumb("CoffeeReg.jpg",75)?>
</td>
I haven't found any of the normal syntax errors, and the code even comes back clean when ran through a syntax checker so that leads me to believe either its a problem with my logic, I'm using the built in functions incorrectly or I have messed up something in the way I am trying to display the new thumbnail only with my lack of exp I don't see where I went wrong. I need to know what I did wrong so I can learn from this mistake.
Okay so after several hours debugging the code that Majid had provided when trying to help me understand where I went wrong with my code I found that the only thing that was needed to make the code he was using work was to add these two lines of code to the end of the secondary php file as shown here...
<?php
$imageUrl = $_GET['file_name'];
$thumbWidth = $_GET['thumb_width'];
// load image and get image size
$mainImage = imagecreatefromjpeg( $imageUrl );
$mainwidth = imagesx( $mainImage );
$mainheight = imagesy( $mainImage );
// calculate thumbnail size
$thumbWidth = intval($mainwidth/4);
$thumbHeight = intval($mainheight/4);;
// create a new temporary image
$tmp_img = imagecreatetruecolor( $thumbWidth, $thumbHeight );
// copy and resize old image into new image
imagecopyresampled( $tmp_img, $mainImage, 0, 0, 0, 0, $thumbWidth, $thumbHeight, $mainwidth, $mainheight );
// display thumbnail
header("Content-type:image/jpeg");
//Also had to remove the second and third argument from this function
imagejpeg( $tmp_img);
//these following two lines here is what was missing
imagedestroy(tmp_img);
imagedestroy(mainImage);
?>
So I guess that one must clear out the image variables of the thumbnail code or it will break it when it runs. Not sure why removing the two arguments from the imagejpeg function was also needed but it was mentioned in my textbook so i tried it, now with these changes this code works. Thank you Majid for all your help and If your post was still here you would be getting full credit with an up vote and best answer.
I'm using this function, to create thumbnails of images uploaded by the user, that I found here: http://webcheatsheet.com/php/create_thumbnail_images.php :
function createThumbs( $pathToImages, $pathToThumbs, $thumbWidth )
{
// open the directory
$dir = opendir( $pathToImages );
// loop through it, looking for any/all JPG files:
if (false !== ($fname = readdir( $dir ))) {
// parse path for the extension
$info = pathinfo($pathToImages . $fname);
// continue only if this is a JPEG image
if ( strtolower($info['extension']) == 'jpg' )
{
echo "Creating thumbnail for {$fname} <br />";
// load image and get image size
$img = imagecreatefromjpeg( "{$pathToImages}{$fname}" );
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresampled( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height );
// save thumbnail into a file
imagejpeg( $tmp_img, "{$pathToThumbs}{$fname}" );
}
}
// close the directory
closedir( $dir );
}
This function works fine and does exactly what I want it to, but, despite this, I'm still getting errors from it. See errors below:
Warning: opendir(images/008/01/0000288988r.jpg,images/008/01/0000288988r.jpg) [<a href='function.opendir'>function.opendir</a>]: The directory name is invalid. (code: 267)
Warning: opendir(images/008/01/0000288988r.jpg) [<a href='function.opendir'>function.opendir</a>]: failed to open dir: No error
Warning: readdir() expects parameter 1 to be resource, boolean given
The problem, I think, is that I'm passing an actual file, not just a directory, into the parameters for the function. This is the case for $pathtoimages and $pathtothumbs. The function is supposed to search through the directory passed to it to find all images with the .jpg extension. But I would like to just perform the function on the one image uploaded at the time of upload. Is there someway to edit this function to allow for this?
Thanks in advance
the $pathToImage must point to image file
remove
$dir = opendir( $pathToImages );
if (false !== ($fname = readdir( $dir ))) {
// parse path for the extension
$info = pathinfo($pathToImages . $fname);
add $info = pathinfo($pathtoImages); // for the file name
$fname = $info['filename']
replace {$pathToImages}{$fname} with $pathToImages only, since its the image file.
btw, this code is not verbose.
function createThumbs( $pathToImages, $pathToThumbs, $thumbWidth )
{
// load image and get image size
$img = imagecreatefromjpeg( "{$pathToImages}" );
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresampled( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height );
// save thumbnail into a file
imagejpeg( $tmp_img, "{$pathToThumbs}" );
}
Think I prematurely posted this question. Thanks for help from everyone.
#csw looks like your solution may have worked, but I got mine working too so I didn't test it.
Quick and dirty:
function createThumb( $pathToImage, $pathToThumb, $thumbWidth )
{
$fname = $pathToImage;
echo "Creating thumbnail for {$fname} <br />";
// load image and get image size
$img = imagecreatefromjpeg( "{$pathToImage}{$fname}" );
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresampled( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height );
// save thumbnail into a file
imagejpeg( $tmp_img, "{$pathToThumb}{$fname}" );
}
You have to use that function like this:
createThumbs("path_to_image", "path_to_thumb", "thumb_width");
Replacing the arguments. Notice the word "path", it's a directory, like "../images/02", and you are using probably the path along with the picture name, like this:
createThumbs("images/008/01/0000288988r.jpg", " ......
And it should be:
createThumbs("images/008/01/" ...
I have encountered a problem with creating a thumbnail from an uploaded image file, then to upload it to the server.
As of now, I have a function that creates the thumbnail, saves it as a temp and returns it to the caller.
Then what I try to do, is upload that created thumb image with move_uploaded_file(tempthumb, path);
Here is the createThumb function and the caller:
function createThumb( $image, $thumbWidth )
{
// load image and get image size
$img = imagecreatefromjpeg( "{$image}" );
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresized( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
return $tmp_img;
}
return $tmp_img; // Here I return the new image. Is this the proper way to get a binary image back??
Here is the caller:
$thumb = createThumb($_FILES['propform-previmg']['tmp_name'], $max_previmg_width);
$filenamepath = $src_dir . '/thumb/' . $_FILES['propform-previmg']['name'];
if ( !move_uploaded_file($thumb, $filenamepath ))
echo "Error moving file {$filenamepath}";
I tried to just upload the uploaded file directly without trying to make a thumbnail first, and that worked fine. So I guess there is some error with the variable I return from the createThumb function, but I can't figure out exactly what.
Also, I need to do the upload from the caller code, and not inside the createThumb function with imagejpeg(file, path).
Thank you!
First line from the manual:
This function checks to ensure that the file designated by filename is a valid upload file (meaning that it was uploaded via PHP's HTTP POST upload mechanism). If the file is valid, it will be moved to the filename given by destination.
Your thumbnail wasn't uploaded, it even isn't a file yet, but just an image resource. To write the image, call something like imagejpeg($resource, $filename) to write it to the path specified in $filename.
I want to upload an image from disk, resize it, and then upload it to Amazon S3.
However, I cant get the proper image output from imagejpeg().
heres my code:
$sourceUrl = $_FILES['path']['tmp_name'];
$thumbWidth = '100';
$thumbid = uniqid();
$img = imagecreatefromjpeg($sourceUrl);
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresampled($tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
// output the image
imagejpeg($tmp_img);
// upload thumbnail to s3
$s3->putObjectFile($tmp_img, "mybucket", $thumbid, S3::ACL_PUBLIC_READ);
Firebug gives me this error :
illegal character
[Break on this error] (�����JFIF���������>CREATOR: g...(using IJG JPEG v62), default quality\n
If I modify imagejpeg this way,
imagejpeg($tmp_img, 'abc.jpg');
then I get the same error. :(
Can i get some help here please ?
If you check the documentation of imagejpeg you can see it outputs the image, it means the way you call it it gets sent to the browser. You can get it to save to a file the second way you call it - by passing a filename in the second parameter.
Also, $tmp_img is an image resource, not a ready-to-use image file.
I don't know how your upload function works, but: if you need the file contents to upload, do it like this:
ob_start();
imagejpeg($tmp_image);
$image_contents = ob_get_clean();
$s3->putObjectFile($image_contents, "mybucket", $thumbid, S3::ACL_PUBLIC_READ);
if you need a filename to upload:
$filename = tempnam(sys_get_temp_dir(), "foo");
imagejpeg($tmp_image, $filename);
$s3->putObjectFile($filename, "mybucket", $thumbid, S3::ACL_PUBLIC_READ);
You have to define the header:
header('Content-type: image/jpeg');
1) $tmp_img is a resource not a file. You probably need to save the image to disc and use that for putObjectFile
2) You probably need to tell S3 that the file you're uploading is of type image/jpeg
Well guys thank you very much again, I screwed around a bit more and combining that with your responses I got this working as follows :)
$thumbid .= ".jpg";
$img = imagecreatefromgif($sourceUrl);
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresampled($tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
$path = '/var/www/1.4/wwwroot/cdn/'.$thumbid;
// output the image
if(imagegif($tmp_img, $path)){
$thumblink = "";
// upload thumbnail to s3
if($s3->putObjectFile($path, "mybucket", $thumbid, S3::ACL_PUBLIC_READ)){
$thumblink = "http://dtzhqabcdscm.cloudfront.net/".$thumbid;
imagedestroy($tmp_img);
}
return $thumblink;
}
I am creating thumbnails of fixed height and width from my PHP script using the following function
/*creates thumbnail of required dimensions*/
function createThumbnailofSize($sourcefilepath,$destdir,$reqwidth,$reqheight,$aspectratio=false)
{
/*
* $sourcefilepath = absolute source file path of jpeg
* $destdir = absolute path of destination directory of thumbnail ending with "/"
*/
$thumbWidth = $reqwidth; /*pixels*/
$filename = split("[/\\]",$sourcefilepath);
$filename = $filename[count($filename)-1];
$thumbnail_path = $destdir.$filename;
$image_file = $sourcefilepath;
$img = imagecreatefromjpeg($image_file);
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
if($aspectratio==true)
{
$new_height = floor( $height * ( $thumbWidth / $width ) );
}
else
{
$new_height = $reqheight;
}
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresized( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height );
// save thumbnail into a file
$returnvalue = imagejpeg($tmp_img,$thumbnail_path);
imagedestroy($img);
return $returnvalue;
}
and I call this function with following parameters
createThumbnailofSize($sourcefilepath,$destdir,48,48,false);
but the problem is the resulting image is of very poor quality, when I perform the same operation with Adobe Photo shop, it performs a good conversion.. why it is so? I am unable to find any quality parameter, through which I change the quality of output image..
Use imagecopyresampled() instead of imagecopyresized().
if it is image quality you are after you need to give the quality parameter when you save the image using imagejpeg($tmp_img,$thumbnail_path,100) //default value is 75
/*creates thumbnail of required dimensions*/
function
createThumbnailofSize($sourcefilepath,$destdir,$reqwidth,$reqheight,$aspectratio=false)
{
/*
* $sourcefilepath = absolute source file path of jpeg
* $destdir = absolute path of destination directory of thumbnail ending with "/"
*/
$thumbWidth = $reqwidth; /*pixels*/
$filename = split("[/\\]",$sourcefilepath);
$filename = $filename[count($filename)-1];
$thumbnail_path = $destdir.$filename;
$image_file = $sourcefilepath;
$img = imagecreatefromjpeg($image_file);
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
if($aspectratio==true)
{
$new_height = floor( $height * ( $thumbWidth / $width ) );
}
else
{
$new_height = $reqheight;
}
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresized( $tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height );
// save thumbnail into a file
$returnvalue = imagejpeg($tmp_img,$thumbnail_path,100);
imagedestroy($img);
return $returnvalue;
}
You could also consider using ImageMagick (http://us3.php.net/manual/en/book.imagick.php) instead of Gd. I had the same problem just a couple of days ago with Java. Going for ImageMagick instead of Java Advanced Images resultet in a huge quality difference.
tried using the php.Thumbnailer ?
$thumb=new Thumbnailer("photo.jpg");
$thumb->thumbSquare(48)->save("thumb.jpg");
Result photo will be 48x48px. Easy right? :)
You might also want to take a look at the Image_Transform PEAR package. It takes care of a lot of the low-level details for you and makes creating and manipulating images painless. It also lets you use either GD or ImageMagick libraries. I've used it with great success on several projects.