Php - 在表格行中显示查询信息[关闭]

Keep getting this same error message

Warning: mysqli_error() expects exactly 1 parameter, 0 given in C:\xampp\htdocs\demetriusdesign777.com\includes\functions.php on line 29 Query FAILED

<?php 

    if(isset($_POST['submit'])) {

        $db['db_host'] = "localhost";
        $db['db_user'] = "root";
        $db['db_pass'] = "";
        $db['db_name'] = "dd777";

        foreach($db as $key => $value) {
        define(strtoupper($key), $value);
        }

        $connection = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);

        $username = $_POST['username'];
        $email = $_POST['e-mail'];
        $phone = $_POST['phone'];
        $info = $_POST['info'];
        $budget = $_POST['budget'];
        $website = $_POST['website'];

        $query = "INSERT INTO userinfo(name, email, phone, website, info, budget) ";
        $query .= "VALUES ('$username', '$email' '$phone', '$info', '$budget', '$website')";

        $result = mysqli_query($connection, $query);

        if(!$result) {
            die('Query FAILED' . mysqli_error());

        } else {

        echo "Record Create"; 

            }
        }

?>

The mysqli_error() call requires a parameter. The parameter should be the connection handle.

if(!$result) {
   die('Query FAILED' . mysqli_error($connection));

This would have told you about the actual error in your query which is the missing comma between '$email' '$phone', in your parameter list section.

$query = "INSERT INTO userinfo
            (name, email, phone, website, info, budget) 
          VALUES 
            ('$username', '$email', '$phone', '$info', '$budget', '$website')";

1. Database connection not right.
2. Query Not correct. Missing , (comma) betwwen '$email' and '$phone'
3. put mysqli_error($connection). Require parameter.

Use This edited code.

<?php 
if(isset($_POST['submit'])) 
{
    $DB_HOST = "localhost";
    $DB_USER = "root";
    $DB_PASS = "";
    $DB_NAME = "dd777";

    $connection = mysqli_connect($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);

    $username = $_POST['username'];
    $email = $_POST['e-mail'];
    $phone = $_POST['phone'];
    $info = $_POST['info'];
    $budget = $_POST['budget'];
    $website = $_POST['website'];

    $query = "INSERT INTO userinfo(name, email, phone, website, info, budget) ";
    $query .= "VALUES ('$username', '$email','$phone', '$info', '$budget', '$website')";

    $result = mysqli_query($connection, $query);

    if(!$result) {
        die('Query FAILED' . mysqli_error($connection));

    } else {

    echo "Record Create"; 

        }
    }   
}