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phploopsfor-loopvariable-declarationfunction-declaration
nevotheless
nevotheless

Reputation: 904

multiple variable assignment causing Undefined variable

i've once again something i don't quite understand. I provide you following code:

<?php
class Helper
{
  static function SelectDateTimeForm($type)
  {
    if($type == 'days')
    {
      $r .= "<select name='something'>";
      for ($x = 1; $x <= 31; $x++) {
        $r .= "<option value='$x'>$x</option>";
      }
      $r .= "</select>";
    }
    return $r;
  }
}
?>

So i wanna simply wanna return to whole select stuff inside the $r variable so that i can access it by calling the SelectDateTimeForm() function.

The Problem now is, that while i am inside the if statement (also try'd it with switch case) the $r variable seems to act somehow crazy. When i leave the if and define the $r directly before the return $r everything seems to work.

So why can't i access or modify the $r variable inside the if? And why am i getting the Undefined variable Notice.

Thanks in advice.

Upvotes: 1

Views: 163

Answers (2)

Osama Jetawe
Osama Jetawe

Reputation: 2705

declare $r before if to be gloabl variable

class Helper
{
  static function SelectDateTimeForm($type)
  {
    $r = '';  
    if($type == 'days')
    {
      $r .= "<select name='something'>";
      for ($x = 1; $x <= 31; $x++) {
        $r .= "<option value='$x'>$x</option>";
      }
      $r .= "</select>";
    }
    return $r;
  }
}

Upvotes: 1

Niranjan N Raju
Niranjan N Raju

Reputation: 11987

Its not because you have multiple assignment, its because you have not declared it. So declare $r above if().

$r = "";
if($type == 'days'){
 .....
}

If $type is not equal to days, it will not enter if(), but you are returning $r which is not declared.

Upvotes: 1

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