cannot display resultant image (php code)

Question

i an trying to get the result(image) from url but there is some thing error in the code my code is:

<?php
if(isset($_POST['submit']))
{
$url= $_POST['url'];
$qrurl= 'http://www.qr-code-generator.com/phpqrcode/getCode.php?
cht=qr&chl='.$url.'&chs=180x180&choe=UTF-8&chld=L|0';
$result = file_get_contents("$qrurl");
echo $result;
}
else
{
echo"<form method='post' action='index.php' >
<input name='url' class='material' placeholder='Enter url...' type='url'autofocus/></br><br>
<input name='submit' id='submit' type='submit' value='submit'/></br><br>
</form>
";
}
?>

the above code displays nothing but when i open the url in browser it displays qr code.

Funk Forty Niner · Accepted Answer · 2017-04-11 15:23:46Z

2

You need a (PNG) header since that is the type of file it is creating, and to make sure the URL is in one line and not broken into two; very important.

<?php 
if(isset($_POST['submit']))
{
$url= $_POST['url'];
$qrurl= 'http://www.qr-code-generator.com/phpqrcode/getCode.php?cht=qr&chl='.$url.'&chs=180x180&choe=UTF-8&chld=L|0';
$result = file_get_contents("$qrurl");
echo $result;

header ('Content-Type: image/png');
exit;

}
else
{
echo"<form method='post' action='index.php' >
<input name='url' class='material' placeholder='Enter url...' type='url'autofocus/></br><br>
<input name='submit' id='submit' type='submit' value='submit'/></br><br>
</form>
";

}

You may also have to add ob_start(); at the top after the opening <?php.

And that the file is not encoded with a byte order mark. This too will throw an error.

The file can be encoded as ANSI or UTF-8 without BOM.

edited Apr 11, 2017 at 15:23

answered Apr 11, 2017 at 15:07

Funk Forty Niner

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2 Comments

Sri Charan Over a year ago

errors:" failed to open stream: Connection timed out" and"Cannot modify header information"

Funk Forty Niner Over a year ago

@SriCharan Go over my answer again; I made a few edits. Follow them completely. This is an encoding issue.

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