PHP停止并显示一行代码

Question

Recently I copied a PHP script to use it for another database. After I correctly edited all the words and links, I got an weird error.

SELECT naam, aantal, prijs 
FROM boeken, bestelling 
WHERE boeken.Boekcode = bestelling.Boekcode 
AND bestelling.Boekcode IN ('101','102') 
AND bestelnummer = 3;

It's not a regular error that say something like

Error on line 42

Basically what the code does, is that when you order books and filled in a form (Name, last name, email etc) it puts that in a database. And afterwards puts it in a "thankyou.html" page.

Here's part of the code that causes this

mysqli_query($con, $query) or die($query . "<br>");
    $bestelnummer = MYSQLI_INSERT_ID($con);

    $object = array_filter($object);
    $objectnaam = join("','",array_keys($object));
    $object = http_build_query($object);
    $object = str_replace('=', ',', $object);
    $object = str_replace('&', "),($bestelnummer,", $object);

    $query = "INSERT INTO bestelling (bestelnummer, Boekcode, aantal) VALUES ($bestelnummer,$object)";
    $result = mysqli_query($con, $query) or die($query."<br>");

    $aantal = "SELECT naam, aantal, prijs FROM boeken, bestelling WHERE boeken.Boekcode = bestelling.Boekcode AND bestelling.Boekcode IN ('".$objectnaam."') AND bestelnummer = $bestelnummer";
    $result = mysqli_query($con, $aantal) or die($aantal . "<br>");

    $res = mysqli_fetch_all($result);
    $prijs = 0;

The $object is an array of the books you choose.

I've Google'd for this problem, and yes, I have Apache and everything enabled.

Sorry if I'm unclear, it's been a while, I can answer any questions you might have.

Answer 1

Well you echo out the query if it fails.

    $aantal = "SELECT naam, aantal, prijs FROM boeken, bestelling WHERE boeken.Boekcode = bestelling.Boekcode AND bestelling.Boekcode IN ('".$objectnaam."') AND bestelnummer = $bestelnummer";
    $result = mysqli_query($con, $aantal) or die($aantal . "<br>");

You might want to check the error message that is returned from the database using mysqli_error().

$result = mysqli_query($con, $aantal) or die(mysqli_error() . "<br>");

Answer 2

first need to refine

$query = "INSERT INTO bestelling (bestelnummer, Boekcode, aantal) VALUES ($bestelnummer,$object)";

you are selecting values insertion for 3 columns and passing only two values

For second query try to use backticks for tablename and column