Question :
I’m making use of a GitHub (xkeshi / image-compressor) project that compresses images using JavaScript. It generates a download of the compressed file with the following link for example blob:http://localhost/945f825f-054a-4170-9d79-ac1dba593d23
Note that the URL starts with blob:
How to upload with PHP from the file to the server, since apparently the file is local?
Data:
GitHub project that originated question: link
What I’ve tried
<?php
//Arquivo que desejo fazer upload blob:http://localhost/945f825f-054a-4170-9d79-ac1dba593d23
$url = 'blob:http://localhost/945f825f-054a-4170-9d79-ac1dba593d23';
$img = '/p/a.jpg';
file_put_contents($img, file_get_contents($url));
?>
Error displaying
Warning:
file_get_contents (blob: link ): failed to open stream: Invalid argument in
C: xampp htdocs 1 a b docs p.php on line 5
Warning: file_put_contents (/p/a.jpg): failed to open stream: No such
file or directory in C: xampp htdocs 1 a b docs p.php on line 5
Answer :
The protocol blob or Object-URLs can only be generated by browser and these URL's can only be accessed / managed in the same instance / browser , which generated them.
Therefore, it is clear that PHP does not have access to URL created.
On the other hand, you can send the file created by the xkeshi/image-compressor library via ajax ( XMLHttpRequest ).
See the example in manual :
new ImageCompressor(file, {
quality: .6,
success(result) {
const formData = new FormData();
formData.append('file', result, result.name);
// Envia a imagem comprimida ao servidor usando XMLHttpRequest.
axios.post('/path/to/upload', formData).then(() => {
console.log('Upload success');
});
},
error(e) {
console.log(e.message);
},
});
Once you’ve done this, you just need to save the uploaded file in PHP, similarly to saving a normal upload.