错误的mysqli语句

Question

我在mysqli的）此错误

mysqli_select_db（期望的是2个参数，1 /home/u751513549/public_html/recent.php给出在第6行

这是密码

<?php // Connects to your Database 
mysqli_connect("localhost", "u751513549_liker", "xxxxxxxx") or die(mysqli_error()); 
mysqli_select_db("u851654599_liker") or die(mysqli_error()); 
$data = mysqli_query("SELECT * 
         FROM token_all 
         ORDER BY RAND() LIMIT 0,9; ") or die(mysqli_error()); 
Print "<table"; 
while ($info = mysqli_fetch_array($data)) { 
    Print "<tr>"; 
    Print " <a href=\"https://www.facebook.com/" . $info['id'] . "/\"/> <img src=\"https://graph.facebook.com/" . $info['id'] . "/picture\"/></a>"; 
} 
Print "</table>"; 

请帮我解决这个错误我是一个初学者。

Answer 1

试试这个（我还添加了另一个参数mysqli_select_db）：

$con = mysqli_connect("localhost", "u751513549_liker", "xxxxxxxx") or die(mysqli_error($con));  
mysqli_select_db($con, "u851654599_liker") or die(mysqli_error($con)); 

Answer 2

第1步：创建连接

$conn = mysqli_connect("localhost","u751513549_liker","xxxxxxxx","u851654599_liker"); 

第2步：运行查询

$query= mysqli_query($conn,"Your Query") or die(mysqli_error($conn)); 

第3步：取记录

$result = mysqli_fetch_all($query); 

步骤4：显示记录

var_dump($result);