giving a error while calling a function in php that variable is undefined

Question

I m using the following code in a PHP script I am trying to call a set of code through function calling(). When I call that function and run that script if shows me the error that the variable $querydigit is undefined.

Can any body tell me how I can call that set of code where I want it.

<?php 
//$querynum = $_SERVER['QUERY_STRING'];

    function calling()
    {
        if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==1)) {$photoname = '1'; }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==2)) {$photoname = '2';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==3)) {$photoname = '3';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==4)) {$photoname = '4';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==5)) {$photoname = '5';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==6)) {$photoname = '6';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==7)) {$photoname = '7';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==8)) {$photoname = '8';  }
    }


if(isset($_GET['1']))
{
$querydigit = 1;
$photoseries = 8;
$foldername = 'founder';
calling();
}

else if(isset($_GET['2']))
{
$querydigit = '2';
$photoseries = 8;
$foldername = 'founder';
calling;
}
}
?>

Define calling to accept an argument and pass the variable to it. Also, read about variable scope. — Felix Kling
– Felix Kling, Commented Feb 11, 2012 at 13:56
Also you can declare the variable as a global inside the function: 'global $querydigit;`. — Stan
– Stan, Commented Feb 11, 2012 at 14:00

alex347 · Accepted Answer · 2012-02-11 13:58:26Z

4

$querydigit is not defined within function's scope. You can fix it in following way:

 function calling($querydigit) {

and then call your function like this:

calling($querydigit);

Here is your code fixed:

   <?php 
//$querynum = $_SERVER['QUERY_STRING'];

    function calling($querydigit)
    {
        if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==1)) {$photoname = '1'; }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==2)) {$photoname = '2';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==3)) {$photoname = '3';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==4)) {$photoname = '4';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==5)) {$photoname = '5';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==6)) {$photoname = '6';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==7)) {$photoname = '7';  }
        else if(isset($_GET[$querydigit]) && ($_GET[$querydigit]==8)) {$photoname = '8';  }
    }


if(isset($_GET['1']))
{
$querydigit = 1;
$photoseries = 8;
$foldername = 'founder';
calling($querydigit);
}

else if(isset($_GET['2']))
{
$querydigit = '2';
$photoseries = 8;
$foldername = 'founder';
calling($querydigit);
}

?>

answered Feb 11, 2012 at 13:58

alex347

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1 Comment

Harish Kumar Over a year ago

thanks for your reply. i have fixed it with calling that function code through a include file also and that works too for me.

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