syntax error, unexpected T_IS_EQUAL [closed]

0
votes
        <?php
    if(isset($_POST['zapisz'])) { 
    $zapytanie_update = mysql_query("UPDATE `emotki` SET `kod` = '".$_POST['kod']."',`opis` = '".$_POST['opis']."', `glowna` = '".$_POST['glowna']."' WHERE id= $id");
    echo 'Emotka zapisana.<br/><a href="index.php">&laquo; Powrót</a>';
    }

    $usun = $_POST['usun'];
    if(isset($usun == 1)
    {
    $usuwaj = mysql_query("DELETE FROM `emotki` WHERE 'id'= $id");
    }



else {
?>

i have ( ! ) Parse error: syntax error, unexpected T_IS_EQUAL, expecting ',' or ')' in C:\wamp\www\emotki_admin\edytuj.php on line 63

line 63 is if(isset($usun == 1)

where is error?

2
php
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. – Kermit

2 Answers

6
votes

You're missing a parenthesis:

if(isset($usun) == 1)

But the comparison is unnecessary anyway. All you need is:

if(isset($usun))

since isset() returns a boolean value.

2
votes

You're missing a parenthesis:

if(isset($usun == 1)

should be

if(isset($usun == 1))