警告：mysqli_select_db（）期望参数1为mysqli，第13行给出的字符串[duplicate]

Question

This question already has an answer here:

• mysqli_select_db() expects parameter 1 to be mysqli, string given 2 answers

<?php //start php tag 
//include connect.php page for database connection
include('connect.php'); 

if(isset($_REQUEST['submit'])!='')
    if ($_REQUEST['name']==''||$_REQUEST['email']=''||$_REQUEST['password']==''||$_REQUEST['repassword']==''){
        Echo "please fill theempty field.";
    }Else{
        $mysqli="insert into user(naam,email,password)values('".$_REQUEST['name']."','".$_REQUEST['email'].",'".$_REQUEST['password']."')";
        $res=mysqli_query($mysqli);
        if($res) { 
            Echo "Recordsuccessfully inserted"; 
        } Else{
            Echo "There is some problem in inserting record"; 
        } 
    } 
}
?>

I keep getting the same error and can't find what's actually the problem, or maybe I just didn't see.

this is my connection

<?php
$hostname="localhost"; 
$username="root"; 
$password="";
$database="website";
$mysqli=mysqli_connect($hostname,$username,$password,$database);
if(! $mysqli){
    die("Connection Failed:".mysqli_error());
}
mysqli_select_db($database,$mysqli);
?>

</div>

Answer 1

You only pass 1 argument into the function. This is wrong. You need two arguments. You have only passed the second.

The first argument is the connection.

http://php.net/manual/en/function.mysqli-connect.php

Please refer to the manual:

bool mysqli_select_db ( mysqli $link , string $dbname )

http://php.net/manual/en/mysqli.select-db.php