Solved

Radio Input throwing Undefined Index

Posted on 2006-07-07
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382 Views
Last Modified: 2008-03-10
Hi,

I have a form that submits values from the following radio button:

<td height="185" align="left" valign="top">
<p><font face="Tahoma" size="2">How many people (on average) will be watching this presentation? <br>
<input type="radio" name="Watching" value="Less than 5">
Less than 5 <br>
<input type="radio" name="Watching" value="6 to 15">
6 to 15<br>
</font><font face="Tahoma" size="2">
<input type="radio" name="Watching" value="16 to 30">
16 to 30 <br>
<input type="radio" name="Watching" value="31 to 100">
31 to 100 <br>
<input type="radio" name="Watching" value="101 to 500">
101 to 500 <br>
<input type="radio" name="Watching" value="More than 500">
More than 500</font></p></td>

I then submit the code to another page where I attempt to verify it:

function cleantext($text)
{
      if( isset( $text )){
            return $text;
      } else {
            $r = "";
            return $r;
      }
}

//Submission conversion
$Watching = cleantext($_POST["Watching"]);


But i get the error:

Notice: Undefined index: Watching in C:\Sites\Single27\bsowards\webroot\presteam\Services\Packaged\3 Templates\formsubmit.php on line 49


I'm using PHP 5 on Windows Server 2003, can anyone tell me how to check if a Radio submitted value is valid? is it because it is an array? Sample code would be much appreciated thanks!

0
Comment
Question by:bsowards
  • 4
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9 Comments
 
LVL 49

Expert Comment

by:Roonaan
ID: 17057900
You can use below code, to debug your request variables:
echo '<pre>'.var_export($_POST,true).'</pre>';

-r-
0
 

Author Comment

by:bsowards
ID: 17057980
your debug did not provide any information or did not work. I'm still getting the error with no additional information.
0
 

Author Comment

by:bsowards
ID: 17057990
whoops! put at end of code so didn't get to it.

It's showing that the value "Watching" isn't even included in the index, which is what I assumed from the error. So how come my detection script isn't handling that?
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LVL 49

Expert Comment

by:Roonaan
ID: 17058074
Are the radio buttons between your <form> and </form> tags?

-r-
0
 
LVL 40

Expert Comment

by:Richard Quadling
ID: 17058076
Change $_POST to $_GET.

This should match the form's action (POST or GET).

Can you show the complete form you are using?

0
 

Author Comment

by:bsowards
ID: 17058113
I don't want to use $_GET, I prefer $_POST.

The form works fine when one of the radio buttons are selected, but doesn't if one isn't. Most sites recommend setting one radio button to the default by putting "checked" on one of the options.

I'd prefer to know how to check if a variable is IN the index, that way I don't have to worry about it. Thanks.
0
 
LVL 49

Expert Comment

by:Roonaan
ID: 17058135
if(isset($_POST['Watching']))

-r-
0
 

Author Comment

by:bsowards
ID: 17058300
Yes well, my function is set to do that, but it doesn't work.

function cleantext($text)
{
     if( isset( $text )){
          return $text;
     } else {
          $r = "";
          return $r;
     }
}

//Submission conversion
$Watching = cleantext($_POST["Watching"]);
0
 
LVL 49

Accepted Solution

by:
Roonaan earned 500 total points
ID: 17058319
No, your function isn't. You cannot pass a non-existing variable and then test if it is set.

So you would use:
$Watching = isset($_POST['Watching']) ? $_POST['Watching'] : '';

Or
$Watching = cleantext($_POST, $Watching)
And use:
function cleantext($array, $key) {
  if(isset($array[$key])) {
  return $array[$key];
  } else { return '';}
}

-r-
}

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