Function parameter

PHP
1

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.


<?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
 
}

?>
2

What about doing this? (note the file extension)


<?
include("../display.php");

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

It’s usually better (more flexible) to return values from functions and echo using the calling code (rather than echoing directly from the function)

3

You have to call the function:


<?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
}

// call the function (note that I put the parameters in another order, if that's not what you want, just change them)
test_display($find, $field, $searching); 
?>
4

Hello

Thanks for responding.

I was trying to keep it simple, below is the function call being used.

I assumed that if I indicate the echo outside the function work, it would imply I am calling the function.

NOTE: The function is a separate script that is linked to the page using


<?
include("../display");  //link separate script

   test_display($searching, $field, $find);//call function
?>
5

So all you get from the echo inside the function is:

(2a), (2b), (2c)

?