CODEWITHSUNDEEP
phpfunctionvariablesundefined
Mentalhead
Mentalhead

Reputation: 1505

Undefined variable problem with PHP function

I'm a PHP newbie, so I have a minor problem functions. I have this line of code:

<?php
$ime=$_POST["ime"];
$prezime=$_POST["prezime"];
$pera="string";
if (empty($ime)||empty($prezime)){
    echo "Ne radi, vrati se nazad i unesi nesto!";
}
function provera($prom){
    if (preg_match("/[0-9\,\.\?\>\.<\"\'\:\;\[\]\}\{\/\!\\\@\#\$\%\^\&\*\(\)\-\_\=\+\`[:space:]]/",$prom)){
        echo "Nepravilan unos imena ili prezimina!";
        echo $pera;
        }
}
provera($ime);
provera($prezime);
?>

Anyway, when I try this code I always get an error message saying that there's a error on on line 11 (the bold part of the code) and no variable is echoed. I'm guessing that it gives me that error because my variable isn't defined inside of that function, but I need to define it outside of the function so is there a way to do this?

Upvotes: 10

Views: 47902

Answers (5)

WINGLEUNG CHOI
WINGLEUNG CHOI

Reputation: 134

If your PHP version is on 5.3 or later versions, closure can be applied.

Closures may also inherit variables from the parent scope.

use is the php syntax to implement closure.

ref: Anonymous functions

    <?php
    // $ime=$_POST["ime"];
    // $prezime=$_POST["prezime"];
    $pera="string";
    $prezime = "Ne radi, vrati se nazad i unesi nesto!";
    // if (empty($ime)||empty($prezime)){
    //     echo "Ne radi, vrati se nazad i unesi nesto!";
    // }
    $provera = function ($prom) use ($pera) {
        if (preg_match("/[0-9\,\.\?\>\.<\"\'\:\;\[\]\}\{\/\!\\\@\#\$\%\^\&\*\(\)\-\_\=\+\`[:space:]]/",$prom)){
            echo "Nepravilan unos imena ili prezimina!";
            echo $pera;
        }
    };

    // $provera($ime);
    $provera($prezime);

Upvotes: 7

John Parker
John Parker

Reputation: 54445

This is because you're using the $pera variable (which exists only in the global scope) inside a function.

See the PHP manual page on variable scope for more information.

You could fix this by adding global $pera; within your function, although this isn't a particularly elegant approach, as global variables are shunned for reasons too detailed to go into here. As such, it would be better to accept $pera as an argument to your function as follows:

function provera($prom, $pera){
    if (preg_match("/[0-9\,\.\?\>\.<\"\'\:\;\[\]\}\{\/\!\\\@\#\$\%\^\&\*\(\)\-\_\=\+\`[:space:]]/",$prom)){
        echo "Nepravilan unos imena ili prezimina!";
        echo $pera;
        }
}

Upvotes: 19

AllisonC
AllisonC

Reputation: 3099

In your function function provera($prom) add a line that says 

global $pera;

Upvotes: 4

JohnP
JohnP

Reputation: 50019

You can't use $pera inside the method like that because it's not defined inside the method scope.

If you want to use the method, pass it as a parameter.

function provera($prom, $pera){ //passed as a param
    if (preg_match("/[0-9\,\.\?\>\.<\"\'\:\;\[\]\}\{\/\!\\\@\#\$\%\^\&\*\(\)\-    \_\=\+\`[:space:]]/",$prom)){
        echo "Nepravilan unos imena ili prezimina!";
    echo $pera;
}

Upvotes: 2

lvictorino
lvictorino

Reputation: 575

It sounds like you have nothing set in your $pera variable. If you have to define a variable outside a function, try passing its value as argument to your function.

function echoMyVar( $myVar )
{
   echo $myVar;
}


$p = "toto";
echoMyVar($p);

Upvotes: 2

Related Questions