我的代码出错，但我无法找到解决方法

Question

Not able to connect to the database. My error:

[Sat Jul 29 23:02:56.747139 2017] [proxy_fcgi:error] [pid 13049:tid 140689614522112] [] AH01071: Got error 'PHP message: PHP Warning: mysqli::mysqli(): (HY000/1045): Access denied for user 'timavnl_1'@'localhost' (using password: YES) in /home/timavnl/domains/tim-av.nl/public_html/6628/job/db_connect.php on line 2 PHP message: PHP Warning: mysqli::prepare(): Couldn't fetch mysqli in /home/timavnl/domains/tim-av.nl/public_html/6628/job/index.php on line 5 PHP message: PHP Warning: main(): Couldn't fetch mysqli in /home/timavnl/domains/tim-av.nl/public_html/6628/job/index.php on line 10 PHP message: PHP Notice: Trying to get property of non-object in /home/timavnl/domains/tim-av.nl/public_html/6628/job/index.php on line 23 PHP message: PHP Fatal error: Call to a member function fetch() on null in /home/timavnl/domains/tim-av.nl/public_html/6628/job/index.php on line 24 '

My DB connect code:

<?php
$db = new mysqli("localhost","timavnl_1","password"."timavnl_1") or 
die("ERROR! With connection");
?>

index.php code: Picture of: index.php

Answer 1

You have a typo mistake. You have given . after password, change this with ,, use like this

"<"?php $db = new mysqli("localhost","timavnl_1","password","timavnl_1") or die("ERROR! With connection"); ?>

The right syntax for make a connection object is, parameters must be sephrated by , operator.

Syntax

$mysqli = new mysqli("localhost", "user", "password", "database");

Note: Depending on the connection function used, assorted parameters can be omitted. If a parameter is not provided, then the extension attempts to use the default values that are set in the PHP configuration file. Check the mannual here