PHP function not working

Question

<?php
$name = $_POST['name'];
$namecheck  = namecheck($name);

    function namecheck($name){

            if(strlen($name)>0)
            {
                return TRUE;
            }
            else if(strlen($name)==0)
            {
                return FALSE;
            }
        };

    function control($namecheck1) 
    {
    if ($namecheck == TRUE) 
    {
        echo "It is TRUE"; 
    } 
    else if ($namecheck == FALSE) 
    {
        echo "It is FALSE";
    }
    };
?>

I wrote that there is no problem in HTML part, there is a problem in my php functions because I am new in PHP. Can you make it proper.

I think you will understand what I want to do in the functions its simple Im trying to do if it is true I want to see "it is TRUE" in the screen. Else .....

Lightness Races in Orbit · Accepted Answer · 2011-05-29 19:11:51Z

6

Take a look at the variables. They don't match:

function control($namecheck1) 
    {
    if ($namecheck == TRUE)

You also never actually invoke that function.

answered May 29, 2011 at 19:11

Lightness Races in Orbit

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Comments

Nanne · Accepted Answer · 2011-05-29 19:11:43Z

2

You're not calling your 'control' function. try starting with

$name = $_POST['name'];
$namecheck  = namecheck($name);
control($namecheck);

Also, your definition of you function is wrong (or the variable you use is). You can change the function to this

 function control($namecheck)

Of the if's to

if ($namecheck1 == TRUE)

in the end the name after control( is the one you should check for in the if's

answered May 29, 2011 at 19:11

Nanne

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Comments

MDEV · Accepted Answer · 2011-05-29 19:12:18Z

1

In your control function, the parameter is called $namecheck1 at first, but you only call it $namecheck when you try to use it inside the function.

answered May 29, 2011 at 19:12

MDEV

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Comments

Haru · Accepted Answer · 2011-05-29 19:13:26Z

1

It looks as if you are not calling control function.

answered May 29, 2011 at 19:13

Haru

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Comments

Vinnyq12 · Accepted Answer · 2011-05-29 19:13:58Z

1

your are referencing $namecheck in the function "control" but the parameter passed is named $namecheck1. $namecheck in the scope of function "control" is undefined.

answered May 29, 2011 at 19:13

Vinnyq12

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Comments

Vytautas · Accepted Answer · 2011-05-29 19:24:25Z

1

Some tips:

instead namecheck() you can use empty()
before using $_POST['name'] you should check if it exists isset() should help

answered May 29, 2011 at 19:24

Vytautas

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Comments

vitthal · Accepted Answer · 2014-03-20 06:43:11Z

0

This works fine

<?php
$name = $_REQUEST['name'];

    function namecheck($name1)
    {

            if(!empty($name1))
            {
                return TRUE;
            }
            else
            {
                return FALSE;
            }
        }


    if (namecheck($name) == TRUE) 
    {
        echo "It is TRUE"; 
    } 
    else if (namecheck($name) == FALSE) 
    {
        echo "It is FALSE";
    }

?>

answered Mar 20, 2014 at 6:43

vitthal

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