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I am trying to work out how to return a base64 result from a server side file and show it in an existing img tag.

I have the following php that is showing the first img ok – but the second one shows as broken.

What is happening here?

<?php

$smiley="data:image/png;base64,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";

if ( isset($_GET['img']) && $_GET['img'] === 'Y' ) {
    echo $smiley;
    exit;
}

echo "<img src='$smiley'> ";


$http=$_SERVER['REQUEST_SCHEME']."://";
$siteurl=$_SERVER['HTTP_HOST'];
$sitepage=$_SERVER['PHP_SELF'];
echo "<img src='$http$siteurl$sitepage?img=Y'>";

2

Answers


  1. Chosen as BEST ANSWER

    Ok - I've solved it:

    if ( isset($_GET['img']) && $_GET['img'] === 'Y' ) {
        $smiley2=str_replace("data:image/png;base64,", "", $smiley);
        header('Content-Type: image/png');
        echo base64_decode($smiley2);
        exit;
    }
    

  2. PHP has support for data URIs:

    if ('Y' === ($_GET['img'] ?? null)) {
        copy($smiley, 'php://output');
        exit;
    }
    

    And you don’t need to send content-type when the context is the src of an image in hypertext.

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