Using php to echo a dynamic URL correctly

Question

I am using php and conditional code to give a dynamic url to a photo. The result should read as http://example.com/biophotos/1.jpg. But instead I am getting http://example.com/%22http://example.com/biophotos/1.jpg%22

How can I force it to just give the one url and without the %22 space on the end?

if ($emresult[0]['photo'] = "y") {
echo '<img class=\"alignright\" src=\"http://example.com/biophotos/' .
$theID . '.jpg" width=\"150\" height=\"150\">';
}
else {
echo 'There is no author photo.';
}

You shouldn't escape double quotes when using single quotes for your echo. — icecub
– icecub, Commented Jul 4, 2015 at 5:21
That did the trick removing the escaping. Thanks! However, it doesn't give me the else "There is no photo" if the condition is not met. Do you see an obvious reason why? $theID is variable capturing the user's ID from a previous page. — JA4677
– JA4677, Commented Jul 4, 2015 at 5:26

slik · Accepted Answer · 2015-07-04 05:22:57Z

1

There is no reason in escaping double quotes when your using single quotes.

echo '<img class="alignright" src="http://example.com/biophotos/' . $theID . '.jpg" width="150" height="150" />';

answered Jul 4, 2015 at 5:22

slik

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Community · Accepted Answer · 2017-05-23 12:29:46Z

0

As much as I agree with @slik, you also might want to look into %22 (double quotes) added to url out of nowhere

Check if magic quotes is on in your php.ini file. You can look into html_entity_encode() to encode the %22 as a slash.

edited May 23, 2017 at 12:29

CommunityBot

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answered Jul 4, 2015 at 5:28

Rohan

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