PHP MySql - Notice: Undefined variable Error

Question

I have this Code For Show selected approved Items in Dropdonn Menu :

PHP:

  if (!isset($form['approved']) || $form['approved'] == '0')
     $approved0 = 'selected';
  elseif (isset($form['approved']) && $form['approved'] == '1')
     $approved1 = 'selected';
  else
     $approved2 = 'selected';

HTML:

<select class="form-control" name="approved" class="formEditSelect">
     <option value="1" ' . $approved1 . '>Active</option>
     <option value="0" ' . $approved0 . '>Inactive</option>
     <option value="2" ' . $approved2 . '>Expired</option>
</select>

Now I see This error:

Notice: Undefined variable: approved1 in C:\xampp\htdocs\test\edit.php on line 207

Notice: Undefined variable: approved2 in C:\xampp\htdocs\test\edit.php on line 207

How do i can fix This error?

exactly and the reason is when the form is not posted `$approved1 and $approved2` is undefined. — Abhik Chakraborty, May 01 '14 at 10:27

Rakesh Sharma · Answer 1 · 2014-05-01T10:32:58.073

initialize variable first with blank or some values

$approved1 = '';

cause your vars in condition so default will be undeclared so initialize them first

or use check with isset() or empty() like :-

$approved1 = (!empty($approved1) ? $approved1 : '');

so your code will be

$approved0 ='';
$approved1 ='';
$approved2 ='';
if (!isset($form['approved']) || $form['approved'] == '0')
     $approved0 = 'selected';
  elseif (isset($form['approved']) && $form['approved'] == '1')
     $approved1 = 'selected';
  else
     $approved2 = 'selected';

score 1 · Accepted Answer · answered May 01 '14 at 10:29

Try This:

$approved0=null;
$approved1=null;
$approved2 = null;

if (!isset($form['approved']) || $form['approved'] == '0')
     $approved0 = 'selected';
elseif (isset($form['approved']) && $form['approved'] == '1')
     $approved1 = 'selected';
else
     $approved2 = 'selected';

score 0 · Answer 3 · answered May 01 '14 at 10:27

0

declare first $approved1 = null; and $approved2 = null; or $approved1 = ''; you should first initialise these variables

answered May 01 '14 at 10:27

hellosheikh

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PHP MySql - Notice: Undefined variable Error

3 Answers3