if ($_SERVER["REQUEST_METHOD"]=="POST") {
$updatedate=$_POST['date'];
$updateday=$_POST['day'];
$updateplace=$_POST['place'];
$updatehighlight=$_POST['highlight'];
$updatediscription=$_POST['discription'];
$sqlupdate="UPDATE $tableselect SET entrydate='$updatedate',day='$updateday',place='$updateplace',highlight='$updatehighlight',discription='$updatediscription' WHERE id ='$getid'";
$sqlquery=mysqli_query($db,$sqlupdate);
if (!mysqli_query($db,$sqlquery)) {
echo "error " .$sqlquery. "<br>" . mysqli_error($db);
}
}
it showing this error:
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '1' at line 1
Real simple, you're running the query function twice, that's why you're getting 1.
So
$sqlquery=mysqli_query($db,$sqlupdate);
if (!mysqli_query($db,$sqlquery)) {...}
needs to be changed to just
$sqlquery=mysqli_query($db,$sqlupdate);
if(!$sqlquery){...}
The first gets executed, and the (if)! operator will also trigger the query function since it was TRUE, as in "(if)not failing".
"For other successful queries mysqli_query() will return TRUE."
Parametrize your query also, you're open to an SQL injection.