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Question about combining query string and a SQL statement

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magicandmight

magicandmight

IS-IT--Management
Aug 27, 2004
77
US
I am using the following procedure to create a query string in the url.

<script type="text/javascript"><!--
function doSomethingSpectacular() {
var splitAtEqual;
var theWholeUrl = document.location;
var splitAtQMark = theWholeUrl.split("?");
if ( isArray( splitAtQMark ) ) {
splitAtEqual = splitAtQMark[1].split("=");
}
if ( isArray( splitAtEqual ) ) {
alert( splitAtEqual[0] + " equals " + splitAtEqual[1] );
}
}
//--></script>

<body onload="doSomethingSpectacular();">

With the link as: <a href="thenextpage.html?mycoolvariable=stuff">Click Me</a>



How would I pull what is after the equal sign into a variable that I can put in the SQL statement:

$query = "SELECT CoName,Anchor FROM list WHERE Gcat LIKE '%worship%' and Anchor <> '' ORDER by CoName";

where what is after the equal sign would take the place of %worship%
 
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You do this server-side using what ever environment you are using. If I was using PHP I would do it like this:
Code: 
<? $server_side_mycoolvariable = '';
 if (isset($_GET['mycoolvariable'])
  $server_side_mycoolvariable = $_GET['mycoolvariable'];
...
 $query = "SELECT CoName,Anchor FROM list WHERE Gcat LIKE '%[COLOR=red]$server_side_mycoolvariable[/color]%' and Anchor <> '' ORDER by CoName";
...
?>
If you are not using PHP, best to ask in the forum appropriate to your code.

Cheers,
Jeff

[tt]Jeff's Page @ Code Couch
[/tt]

What is Javascript? FAQ216-6094
 
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I am using PHP. Ill try it out. One question though what is the "..." for in the code? Guessing it means the rest of my code.
 
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It didnt like the line that has:
$server_side_mycoolvariable = $_GET['mycoolvariable'];

gave me an error:


Parse error: parse error, unexpected T_VARIABLE
 
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Yes, it is a PHP problem, but let's just solve it quickly... Since we ventured into PHP inadvertly anyway. Please check the code you're given for possible errors of oversight:
Code: 
 if (isset($_GET['mycoolvariable'])[red][b])[/b][/red]
There was a missing parenthesis there. Try to match them up.
 
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Ah... that was my fault... sorry for not making it clear that I was unable to test the code snippet I posted.

And yes, '...' implied "your other code goes here" - but you figured that out :)

Hope you got it running ok!

CHeers,
Jeff

[tt]Jeff's Page @ Code Couch
[/tt]

What is Javascript? FAQ216-6094
 
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