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Re: Forms in Php

Posted by Jerry Stuckle on 10/18/05 17:38

onefastmustang wrote:
> I am using that exact syntax.. I misspoke before.. POST does not work
> either.. The reason I was using GET is that the client wants to be
> able to use the back button on the browser without getting the PAGE
> EXPIRED page..
>
> my form looks like this..
> <input type="checkbox" value= "1" name="roles[]" >
> <input type="checkbox" value= "2" name="roles[]" >
>
> The reciveing code looks like this..
>
>
> $roles = $_GET[roles]. $_POST[roles];
>
>
> if ($roles) {
> $sqls = $sqls. " and profiles.role IN (";
> foreach ($roles as $key => $value){
> $sqls = $sqls. "$value,";
> }
> $sqls = substr("$sqls", 0 , -1);
> $sqls .= ")";
> }
>

Well, $roles won't be an array at this time - you did a string
concatenation on it.

If you're using GET method, use $_GET. If you're using $_POST, use
$_POST. You can also use $_REQUEST to handle either - but personally I
don't like that as well. But that's just my personal preference.

Also, if neither checkbox is checked, $roles will be empty.

A couple of changes:

$roles = isset($_GET['roles']) ? $_GET($roles) : null; // Note quotes!

if (isset($roles) { // Has one been set?
if (isarray($roles) { // Ensure it's an array
$sqls = $sqls. " and profiles.role IN (";
foreach ($roles as $key => $value){
$sqls = $sqls. "$value,";
}
$sqls = substr("$sqls", 0 , -1);
$sqls .= ")";
}
else { // Set but not an array
$sqls = $sqls . " and profiles.role = " . $roles . ")"
}
}

(note - not checked for syntax)
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================

 

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