How to display mysqli_query errors

Question

I am trying to display mysqli_error in my php script, if the query fails, I keep getting a black page, but on success, a response is returned, if the query fails, both mysqli_error($conn) and mysqli_errno($conn) Just shows a black page. This is my database connection script

<?php
$ser = "test";
$user = "test";
$pass = "test";
$db="test";
$conn = mysqli_connect($ser, $user, $pass, $db);
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
  }
?>

My PHP Query

<?php
include 'conn.php';
if ($conn) {
    $sql = "INSERT INTO tbl_users (userId, email, pass) VALUES ('$userId', '$email', '$pass')";
    if (mysqli_query($conn, $sql)) {
        echo json_encode(array(
            "status" => "Ok",
            "message" => "Success",
        ));
    } else {
        echo json_encode(array(
            "status" => "Error",
            "message" => mysqli_error($conn),
        ));
    }
}

If the query runs correctly, I am able to get a successful response, but if I use ini_set('display_errors', 1); once an error occurs, I get a blank page;< /strong> I get full page errors but I just want mysqli_error or mysqli_errno.

Answer 1

As of PHP 8.1, the default setting for mysqli error reporting is MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT, which causes errors to throw exceptions instead of just returning false.

If you want to check for errors in your code instead of getting exceptions, use

mysqli_report(MYSQLI_REPORT_OFF);

A better solution is to use exception handlers.

if ($conn) {
    $sql = "INSERT INTO tbl_users (userId, email, pass) VALUES ('$userId', '$email', '$pass')";
    try {
        mysqli_query($conn, $sql);
        echo json_encode(array(
            "status" => "Ok",
            "message" => "Success",
        ));
    catch (mysqli_sql_exception $e) {
        echo json_encode(array(
            "status" => "Error",
            "message" => $e->getMessage()
        ));
    }
}