PHP Notice: Undefined index: status on line

Question

I have this code :

if (isset($form['status']) && $form['status'] == 0)
    $status0 = ' selected';
elseif (isset($form['status']) && $form['status'] == 1)
    $status1 = ' selected';

HTML:

<option <?PHP echo $status0; ?>>YES</option>

Now, I see This Error:

PHP Notice:  Undefined variable: status0 in C:\xampp\htdocs\project\cms.php on line 212

How Do i can fix This?!

It doesn't really matter what value in the MySQL database is. The error indicates that when the variable `$status0` is referenced, it has not yet been defined. (We conclude that the `if` condition did not evaluate to TRUE, and that `$status0` isn't defined anywhere else before line 212. The "fix" is to define the variable somewhere other than just following an `if`. — spencer7593, Apr 19 '14 at 15:23

score 3 · Accepted Answer · answered Apr 19 '14 at 15:13

3

PHP gave you the answer:

PHP Notice:  Undefined variable: status0 in C:\xampp\htdocs\project\cms.php on line 212

Declare your vars before the if statement: $status0 = null;

answered Apr 19 '14 at 15:13

toesslab

5,092
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score 2 · Answer 2 · answered Apr 19 '14 at 15:11

2

It seems that $status0 is not defined, because it skips all your if/elseif blocks. You can fix it by assigning $status0 to the default value before the clauses or add else clause and assign $status0 in it.

answered Apr 19 '14 at 15:11

potashin

44,205
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PHP Notice: Undefined index: status on line

2 Answers2