Uploaded ZIP archive. Convert only PNG to JPEG in PHP - php

For the life of me, I cant figure out how to code part of this process:
Ive Already Completed These Steps:
1. Upload ZIP archive (containing only photos in gif, png, and jpg)
2. Unpack to folder
3. Scan folder for filenames + file extentions
I Need Help With:
4. Convert only PNG to JPG
Any help would be appreciated!
Brandon
EDIT:
Does this make sense?
$directory = "../images/ilike/goldfish/";
$images = glob($directory . "*.jpg");
foreach($images as $image)
{
$pic = imagecreatefrompng($directory);
$bg = imagecreatetruecolor(imagesx($image), imagesy($image));
imagefill($bg, 0, 0, imagecolorallocate($bg, 255, 255, 255));
imagealphablending($bg, TRUE);
imagecopy($bg, $pic, 0, 0, 0, 0, imagesx($image), imagesy($image))
imagedestroy($image);
imagejpeg($bg, $image . ".jpg", 100);
ImageDestroy($bg);
}

Take in the file from a form as usual, open it with http://www.php.net/manual/en/class.ziparchive.php and use http://www.php.net/manual/en/ziparchive.extractto.php to extract the files to an empty folder. Then using the PHP standard file handling functions scan filenames that end with .jpg. Load them in with http://www.php.net/manual/en/function.imagecreatefromjpeg.php and then save them out with http://www.php.net/manual/en/function.imagepng.php .

//here we create directory
----------------------------
$new_folder=mkdir('C:\\wamp\\www\\TestImage\\uploads\\'.$folder_name, 0777, true);
$path="uploads\\".$folder_name."\\";
//uploading files
-----------------------------
$fileName = $_FILES["upload_file"]["name"]; // The file name
$fileTmpLoc = $_FILES["upload_file"]["tmp_name"]; // File in the PHP tmp folder
$fileType = #$_FILES["upload_file"]["application/zip"]; // The type of file it is
$fileSize = $_FILES["upload_file"]["size"]; // File size in bytes
$fileErrorMsg = $_FILES["upload_file"]["error"]; // 0 = false | 1 = true
$kaboom = explode(".",$_FILES["upload_file"]["name"]); // Split file name into an array using the dot
$fileExt = end($kaboom); // Now target the last array element to get the file extension
if (!preg_match("/.(zip)$/i", $fileName) ) {
// This condition is only if you wish to allow uploading of specific file types
//echo "ERROR: Your file was not .zip file";
echo "Please select image files.Supported format are .Zip</br>";
unlink($fileTmpLoc); // Remove the uploaded file from the PHP temp folder
#exit();
}
//unzip to specific location
--------------------------------
if(preg_match("/.(zip)$/i", $fileName))
{
$moveResult= move_uploaded_file($fileTmpLoc, $fileName);
if($moveResult == true)
{
$zip = new ZipArchive;
$res = $zip->open($fileName);
if($res==TRUE)
{
$zip->extractTo($path.$fileName);
echo "<pre>";
print_r($zip);
$zip->close();
} else {
echo 'failed';
}
}
unlink($fileName); // Remove the uploaded file from the PHP temp folder
//exit();
}
//function for Convert only PNG to JPG
$image = imagecreatefrompng($originalFile);
imagejpeg($image, $outputFile, $quality);//$outputFile->define name of output file and $quality is a number between 0 (best compression) and 100 (best quality)
imagedestroy($image);
note->refer for <gdlib> http://www.php.net/manual/en/function.imagejpeg.php

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File is writing into wrong path using PHP

I am uploading file into folder using PHP but my issue is its at a time writing file into 2 different path. My code is below.
if(array_key_exists('pimage',$_FILES)){
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Here I need to write file into PT_USERS_IMAGES_UPLOAD path but before uploading into this path also the file is uploading into project's root path. Here I need to upload only in PT_USERS_IMAGES_UPLOAD path not in project's root path.
It is likely because of these lines:
imagepng(imagecreatefromgif($_FILES['pimage']['tmp_name']), $saveFile);
If you look at the documentation regarding this imagepng(), it will either output an image to the browser (with a proper header) or save the file to disk when you fill out the second parameter, in your case you have used the to parameter ($saveFile). So, once you save it there, you then save it again using the move_uploaded_file($tempFile, $targetFile); which is the one saving it to the proper location.
If you are trying to convert something to PNG, then just do the imagepng() line and remove the move_uploaded_file() line. Change $saveFile to T_USERS_IMAGES_UPLOAD and then you should only get one saved file. Either way, remove one of those methods for saving the file to disk.

PHP upload image different directory

i have a problem with uploading images into different directory.
$path = "../uploads/";
$path2 = "../uploads2/";
$imagename = $_FILES['photoimg']['name'];
$actual_image_name = $imagename;
$uploadedfile = $_FILES['photoimg']['tmp_name'];
$widthArray = array(600,240); //resize width.
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i want to upload image into uploads and uploads2 folders also?
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what's wrong with my code?
After moving the file with move_uploaded_file it isn't available in the location stored in $uploadedfile anymore. For the second file you have to use copy function.
Please try the following:
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Remove this line. I don't know for what purpose it's there.
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To resize the uploaded image use any library to resize it then pass it.
But here is the full code to upload in different directory. But you will have to resize these two $file1 & $file2 to your expected resize file and replace the same in $file1 & $file2
To resize you can use any code suggested here
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$path2 = "../uploads2/";
$file1= $_FILES['photoimg'];
$file2= $_FILES['photoimg'];
$file1_imagename = $file1['name'];
$file2_imagename = $file2['name'];
$file1_actual_image_name = $file1_imagename;
$file2_actual_image_name = $file2_imagename;
$file1_uploadedfile = $file1['tmp_name'];
$file2_uploadedfile = $file2['tmp_name'];
$widthArray = array(600, 240); //resize width.
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php imagecopyresampled - image is empty (all black) after saving it

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Get the solution for .png images

Can not resize multiple images in PHP

I am trying to upload various images into a dynamically created folder on my server, then take each image and resize it while uploading it into the folder as well creating a new image and a new name.. example: image.jpg (original image) and image-resized.jpg (being the thumbnail image).
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Why are you defining the function resizeImage in the for loop? It is being redefined every time the loop iterates. This could be part of the problem. Define the function outside the loop and see if that works.
you can try;
when your first image proccess end use imagedestroy() then second image will proccessed.

PHP: Making thumbnail, why do i get these errors?

When i made this function:
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Don't want to copy it to a separate directory?
Want to just display the image to the user?
Follow the following steps:
2. Uncomment this code:
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/* UNCOMMENT THIS IF YOU WANT */
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Then it stopped working, but outside a function, it works good.
As you can see i added "echo $file...." because i wanted to see if they have value, and they do have the right values.
I just get the error Error: Please select a file to upload.
This function is running after an normal upload image script(full size).
When i call the function i do:
makeThumbnail($_FILES[$fieldname]['type'], $_FILES[$fieldname]['name'], $_FILES[$fieldname]['size'], $_FILES[$fieldname]['tmp_name'], 100);
At my other file where its not in a function, theres no difference only that the variables is:
$file_type = $_FILES['file']['type'];
$file_name = $_FILES['file']['name'];
$file_size = $_FILES['file']['size'];
$file_tmp = $_FILES['file']['tmp_name'];
But it should work, I cant find anything wrong, but it doesnt and i keep getting that error. If i remove the is_uploaded_file function, i get a bunch of another errors.
Make sure you are not using move_uploaded_file() before calling the function.
I use timthumb to process the image into a thumbnail when it outputs it to screen, instead of when it's uploaded.
It means you only have one file and not one master size and one thumb size. TimThumb reduces the size of the file on serverside so it appears nice and smooth on the browserside. Have a look at it: TimThumb Link

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