multiple variable assignment causing Undefined variable

Question

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.

Niranjan N Raju · Accepted Answer · 2015-11-05 10:00:43Z

1

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.

answered Nov 5, 2015 at 10:00

Niranjan N Raju

12k4 gold badges24 silver badges41 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

nevotheless Over a year ago

Yeah i checked my $type variable and it was capitalized... Thanks for giving me the bump!

Osama Jetawe · Accepted Answer · 2015-11-05 10:02:09Z

1

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;
  }
}

answered Nov 5, 2015 at 10:02

Osama Jetawe

2,7057 gold badges27 silver badges45 bronze badges

1 Comment

nevotheless Over a year ago

Okay, i did declare it, the notice disappears but the variable still does not change inside the if statement. What am i missing?

Collectives™ on Stack Overflow

multiple variable assignment causing Undefined variable

2 Answers 2

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related