Snippet call parameter values being stripped

14 Posts

netpuppy Reply #1, 16 years, 11 months ago

I’m trying to pass a url with query as a parameter in a snippet call but it is being modifed before I recieve it in the actual snippet.

Example snippet call:

[!mysnippet? &url=`http://www.example.com/mysite/Show?pid=4&mid=8`!]

The &url parameter recieved in the snippet:

http://www.example.com/mysite/Showpi

The ? is stripped out and the string is terminated 1 character before the = sign.

Is it possible to pass through a string of text like this in a snippet call without it being modified?

Any help would be greatly appreciated.

24,544 Posts

BobRay Reply #2, 16 years, 11 months ago

The parser doesn’t like question marks, ampersands and equals signs in parameters.

One solution is to encode them in the tag and then decode them in the snippet:

[

!mysnippet? &url=`http://www.example.com/mysite/ShowQUESTIONpidEQUALS4AMPERSANDmidEQUALS8`!]

Then in the snippet:

<?php
$search = array('QUESTION','AMPERSAND','EQUALS');
$replace = array('?','&','=');

$url = str_replace($search,$replace,$url);
?>

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14 Posts

netpuppy Reply #3, 16 years, 11 months ago

Thanks BobRay, that’s a nice solution.

I messed around with it myself and put the query section of the url in another parameter.

Snippet Call:

[!mysnippet? &url=`http://www.example.com/mysite/Show` &query=`pid|4, mid|8`!]

Snippet Code:

if(trim($query) != "") {
   $url .= "?";
   $query_array = explode(',',$query);
   for ($q=0;$q<count($query_array);$q++) {
      $qtemp = explode("|",$query_array[$q]);
      if($q != 0) {$url .= "&";}
      $url .= trim($qtemp[0])."=".trim($qtemp[1]);
   }
}

Not as elegant as your solution but hey... it works for me.

24,544 Posts

BobRay Reply #4, 16 years, 11 months ago

Glad you got it working.

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