Why am I getting this parse error in a PHP if statement?

Question

I am getting this error:

Parse error: syntax error, unexpected '!=' (T_IS_NOT_EQUAL) in C:\xampp\htdocs\assurance\confirmation.php on line 131

Here is lines 131-134 of my code:

if ($_POST['password']) != ($_POST['confirm'])  { 
echo '<p class="error">Your passwords do not match.</p>';
$okay = FALSE;
}

Remove the extra parentheses. Your code should be: `if ($_POST['password'] != $_POST['confirm'])`. — Amal Murali, Feb 15 '14 at 16:01
Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. — James, Feb 15 '14 at 16:01

score 1 · Accepted Answer · answered Feb 15 '14 at 16:03

1

if ($_POST['password'] != $_POST['confirm'])  { 
echo '<p class="error">Your passwords do not match.</p>';
$okay = FALSE;
}

the syntax of the if statement is

if (expr)
  statement

answered Feb 15 '14 at 16:03

Aris

score 1 · Answer 2 · answered Feb 15 '14 at 16:03

1

use this

if ($_POST['password'] != $_POST['confirm'])  { 
    echo '<p class="error">Your passwords do not match.</p>';
    $okay = FALSE;
}

answered Feb 15 '14 at 16:03

Dima

2 Answers2