Undefined index in PHP [duplicate]

Answer 1

It should be a notice and not an error.

To fix is you'll have to check whether $_POST['submit'] is set:

if(!isset($_POST['submit'])) {
    ...
}

Answer 2

Options:

• Disable warning messages, by editing the setting in PHP.ini
• Add an @ sign in front of the variable name, which will suppress the error on that particular line.
• Change your code to use isset($_POST['SUBMIT']) before checking it further.

Of these, the third option is definitely the best. You shouldn't assume that any variable supplied by the user will be set as you expect; you should always check that it's set at all, and also that it's set to the expected values. Otherwise you are liable to be open to hacking attacks.

Answer 3

It's where you test to see that is isn't there. It should be !isset($_POST['SUBMIT']). This is because the index, SUBMIT, won't be set, thus won't have a value such as true to pass the if(...). isset() checks to see if the index/variable is actually set.

Try this:

<?php
    if (!isset($_POST['SUBMIT'])){   //ERROR: Undefined index
?>
<H2>Add Employee</H2>
<form action="<?php print $_SERVER['PHP_SELF']; ?>" method="POST">
<table width="400" border="0" cellspacing="1" cellpadding="2">
   <tr>
        <td width="100">SSN</td>
        <td><input name="SSN" type="text" id="SSN"></td>
   </tr>
   <tr>
        <td width="100">&nbsp;</td>
        <td><input name="SUBMIT" type="SUBMIT" id="ADD" value="ADD"></td>
   </tr>
</table>
</form>
<?php
    }
    else {
    //code here
    }
?>

Answer 4

<?php
  if (!isset($_POST['SUBMIT'])){   //ERROR: Undefined index
?>

This tests, if the index is set