Return from function not working

[DragonHighLord]

I have a file that is included. In that file is a function and I need it to return a value to the main page, but this is not working.

This is a simple call to check the number of logins for a user, I put in the echo statements for error checking (to see what is going on)

On main page:
include_once("functions.php");

check_logins();

echo "<br>outside function: " .$logins;

On functions.php
function check_logins()
{
$check=db_query(" select logins from users where name='{$logged["name"]}' ");
$logins=db_result($check,0,0);

echo "<br>inside function: " .$logins;

return $logins;
}

(BTW: I have predefined functions of what db_query and db_result are, so the database queries are working correctly)

On my screen I get this:

inside function: 5
outside function:

There is no value being returned
I have tried:

return $logins;
return ($logins); //space after return
return($logins); //no space

None of this seems to work, anything I'm doing wrong?

[AQHost]

Hi,

Everything is ok except that the $logins variable that you create inside the function is not global and therefore scope is limited to that function only. What you need to do is this:

include_once("functions.php"); 
$logincount = check_logins(); 
echo "<br>outside function: " .$logincount;

return($logins); // correct syntax

Best wishes,
Simon.

[Kerry_Kobashi]

Simply set a variable assigned to check_logins():

On main page:
include_once("functions.php");

$retVal = check_logins();

echo "<br>outside function: " .$retVal;

Call it $retVal, $logins, whatever....

Kerry Kobashi
Kobashi Computing

[devinemke]

another way is to recode your function like this:

function check_logins() 
{ 
$check=db_query(" select logins from users where name='{$logged["name"]}' "); 
$logins=db_result($check,0,0); 
$GLOBALS['logins'] = $logins; 
} 

then the $logins variable will be available in the global scope