⇢ ОП
Ольга Павлова
You have an error in your SQL syntax; check ...MySQL server version for the right syntax to use near '' at line 1
Помогите найти ошибку в php коде
if(empty($_POST['a1'])){
$a1 = 0 ;}
else{
$a1 = 1;
}
if(empty($_POST['a2'])){
$a2 = 0 ;}
else{
$a2 = $_POST['a2'];
}
if(empty($_POST['a3'])){
$a3 = 0 ;}
else{
$a3 = $_POST['a3'];
}
if(empty($_POST['a4'])){
$a4 = 0 ;}
else{
$a4 = $_POST['a4'];
}
if(empty($_POST['a5'])){
$a5 = 0 ;}
else{
$a5 = $_POST['a5'];
}
if(empty($_POST['a6'])){
$a6 = 0 ;}
else{
$a6 = $_POST['a6'];
}
$sql = "UPDATE users SET a1 = $a1, staj = 1, a2 = $a2, a3 = $a3, a4 = $a4, a5 = $a5, a6 = $a6 WHERE login = .$login.";
$query = mysql_query($sql);
if(!$query){
die('updating error'. mysql_error( ));
}
?>