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Function parameter

Posted: Sun Oct 17, 2010 2:05 pm
by assgar
Hello

I am experiencing some issues with a function and I cannot identify the problem.
The function is not receiving any parameter.

The URL values are passed correctly. I used echo to display the parameters outside the function and that worked.
I also used echo to display the parameters inside the function and no parameter were displayed.
So the function is not receiving parameters.

Code: Select all

<?php

$find = $_REQUEST['u_find'];
$field = $_REQUEST['u_field'];
$searching = $_REQUEST['u_search'];


 echo"(1a)$find, (1b)$field, (1c)$searching";//outside function

function test_display($searching, $field, $find)
{
   echo"(2a)$find, (2b)$field, (2c)$searching";//inside function
 
}

?>

Re: Function parameter

Posted: Sun Oct 17, 2010 6:02 pm
by TonsOfFun
assgar wrote:

Code: Select all

<?php

$find = $_REQUEST['u_find'];
$field = $_REQUEST['u_field'];
$searching = $_REQUEST['u_search'];


 echo"(1a)$find, (1b)$field, (1c)$searching";//outside function

function test_display($searching, $field, $find)
{
   echo"(2a)$find, (2b)$field, (2c)$searching";//inside function
 
}

?>
Is this your entire code? If so, then you aren't actually calling the function, only defining it.

Add:

Code: Select all

<?php

test_display($searching, $field, $find);

?>