[SOLVED] $_GET not working as expected when getting %Beach%

[brynkaufman]

I have the following URL.  test.php?test=%Beach%

 

Then one line of code.  echo ($_REQUEST['test']);

 

and the result is.  ¾ach%

 

Why is the %B being read as 3/4 by the $_GET?

 

Thanks.

[Barand]

% in a url followed by 2 hex chars is used for encoding characters.

 

I think you'll find BE is ascii value (in hex) for 3/4

 

%20 gives a space character

[kenrbnsn]

That's because the "%" character followed by another 2 characters is interpreted as an urlencoded character when it's on the URL and PHP is receiving the decoded value. You can reencode it:

<?php
$test = urlencode($_GET['test']);
?>

 

Ken

[Barand]

You need to encode it before putting it into the url.

 

Test script:

<?php
if (isset($_GET['test']))
    echo $_GET['test'], '<hr>';
?>

<a href='?test=%beach%'>TEST 1</a>
<br>

<?php
$test = urlencode('%beach%');
echo "<a href='?test=$test'>TEST 2 </a>";
?>

[brynkaufman]

Sorry, I should have mentioned I tried to urlencode and this is what I get.

 

echo urlencode($_GET['test']);

 

gives %BEach%25 

 

Notice the 25 after the last percent sign.  I am trying to pass this search parementer - %Beach% - for MySQL through the URL so I can't have anything after the % or it won't find the right results.

 

Also, the % is not intended to be a space, it is the wildcard for the search so anything with the word Beach in it will come back in the results.

[kenrbnsn]

Use a different character in the URL and replace it with a "%" in your script.

 

Ken

[brynkaufman]

Thanks Ken,

 

I will just use * as my wildcard and change it to a % in the script.  This seems to work fine.  I thought there might be some function to get the % working but using a * for a wildcard will be even more familiar to my users.