PHP Data-URI to file

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I have a data URI I am getting from javascript and trying to save via php. I use the following code which gives a apparently corrupt image file:

  $data = $_POST['logoImage'];

  $uri = substr($data,strpos($data,",")+1);

  file_put_contents($_POST['logoFilename'], base64_decode($uri));



data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAoAAAAKCAYAAACNMs 9AAAACXBIWXMAAAsTAAALEwEAmpwYAAAAxklEQVQYlYWQMW7CUBBE33yITYUUmwbOkBtEcgUlTa7COXIVV5RUkXKC5AxU EdyZVD4kyKxkwIrr9vd0c7Oih aopinLNsF6Qkg2XW4XJ7LGFsAAcTV6lF5/jLdbALA9XDAXYfthFQVx OrmqKYK88/7rbbMFksALieTnzu9wDYTj6f70PKsp2kwAiSvjXNcvkWpAfNZkzWa/5a9yT7fdoX7rrB7hYh2fXo9HdjPYQZu3MIU8bYIlW20y0RUlXG2Kpv/vfwLxhTaSQwWqwhAAAAAElFTkSuQmCC

Below the code is the actual image as a Data-URI. 'logoImage' is the string above, and $uri is the string minus 'image/jpeg;base64,'.

Answers

A quick look at the PHP manual yields the following:

If you want to save data that is derived from a Javascript canvas.toDataURL() function, you have to convert blanks into plusses. If you do not do that, the decoded data is corrupted:
$encodedData = str_replace(' ','+',$encodedData);
$decodedData = base64_decode($encodedData);

Tuesday, August 23, 2022

dukeofmarmalade-955c5c5aff0c

Thank you all for the answers. I was calling the download as a function in a file with other functions in it so in the end I had to write a script alone, apart from other files. My problem was that I needed it to be safe and to only download a file if it belonged to the user and the user was logged in, so I send all the data I need, ciphered and inside the script I use a series of things to see if the owner is really the logged user. So if anyone wants to know this is the code I used and works perfectly.

<?php
session_start();
$a=$_GET['a'];
$parts=explode("-",$a);
$tres=$parts[0];
$nombre=$partes[1];
$dbcodletra=substr($tres,2);
if($dbcod!=$_SESSION["username"])$boolt=0;
if($ext==1) $extl=".jpg";
if($ext==2) $extl=".jpeg";
if($ext==3) $extl=".tif";
if($ext==4) $extl=".tiff";
if($ext==5) $extl=".pdf";
if($ext==6) $extl=".doc";
if($ext==7) $extl=".docx";
if($tipoproy==1) $dir="rute/".$dbcodletra."/".$nombre.$extl;
if($tipoproy==2) $dir="rute/".$dbcodletra."/".$nombre.$extl;
if($tipoproy==3) $dir="rute/".$dbcodletra."/".$nombre.$extl;
if($tipoproy==4) $dir="rute/".$dbcodletra."/".$nombre.$extl;
if (file_exists($dir) && $boolt) {
    header('Content-Description: File Transfer');
    header('Content-Type: application/octet-stream');
    header('Content-Disposition: attachment; filename='.basename($dir));
    header('Content-Transfer-Encoding: binary');
    header('Expires: 0');
    header('Cache-Control: must-revalidate, post-check=0, pre-check=0');
    header('Pragma: public');
    header('Content-Length: ' . filesize($dir));
    ob_clean();
    flush();
    readfile($dir);
    exit;
}else echo "<meta http-equiv="Refresh" content="0;url=misdocumentos.php">";
?>

Wednesday, August 31, 2022

veeti

This should do it in Python:

import base64
encoded = base64.b64encode(open("filename.png", "rb").read())

Wednesday, December 21, 2022

marionmaiden

What you want is...

new File(uri.getPath());

... and not...

new File(uri.toString());

Note: uri.toString() returns a String in the format: "file:///mnt/sdcard/myPicture.jpg", whereas uri.getPath() returns a String in the format: "/mnt/sdcard/myPicture.jpg".

Saturday, October 29, 2022

user79755

I don't think you will gain much... and if it is a file image, the browser can cache it. I wouldn't bother doing it with CSS unless you have a real need for it.

Friday, November 4, 2022

zen_pak

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PHP Data-URI to file

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