PHP output displays garbage

Question

I've seen a few cases of this online but every solution I try yields the same result. I have tried various versions of this same bit of code but I still am unable to display an image, either from the internet or from a folder.

My code looks like this:

    <?php
    $img = 'anything.jpg';
    $getInfo = getimagesize($img);
    header('Content-type: ' . $getInfo['mime']);
    readfile($img);
    ?>

And then this in the appropriate spot in the html document:

<?php
echo "<img src ='$img'>";
?>

My output looks like this: (only it is about 100 lines long, covering most of the page)

„•J€m’¥Id„©,ƒð‹"È!dY*   
d¨E„Œœ6TêÞ BTç–dáfÎ´§ÝfÎ°É¾,ÙÔÌxà5IõcÁ=£TM´® ‡iNt4ÖMYí;ÅU©ª’§ª7’Ê:¡ú§šß”,¾5™ˆ

The image displays after all this. I've been working at this quite a while now, any help would make a man's day.

Community · Accepted Answer · 2023-11-14 22:25:35Z

3

The content-type of the image is in the array under mime, not img.

header('Content-type: ' . $getInfo['mime']);

More info is available in the PHP docs.

edited Nov 14, 2023 at 22:25

CommunityBot

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answered Feb 26, 2014 at 2:07

Mark Parnell

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2 Comments

12th Over a year ago

Thank you very much for your response. I fixed that. I was hopeful but unfortunately it didn't solve my error.

Mark Parnell Over a year ago

Your code to output the image is all that's in that file, right? And the HTML document is separate? Where in that page is $img defined? Is it just the URL of the first script?

12th · Accepted Answer · 2014-02-26 19:06:50Z

0

I ended up solving my problem using the following code:

<?php
$img = 'anything.png';
header("Content-type:image/png");
imagepng($img);
?>

<?php
echo "<img src ='$img'>";
?>

answered Feb 26, 2014 at 19:06

12th

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