I don't find any error.this code doing well.up

问题:

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

回答1:

Real simple, you're running the query function twice, that's why you're getting 1.

• 1 being boolean TRUE.
• (MySQL) Boolean Literals

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".

• http://php.net/manual/en/mysqli.query.php

"For other successful queries mysqli_query() will return TRUE."

Parametrize your query also, you're open to an SQL injection.

• https://en.wikipedia.org/wiki/Prepared_statement