錯誤的mysqli語句

Question

我在mysqli的）此錯誤

mysqli_select_db（期望的是2個參數，1 /home/u751513549/public_html/recent.php給出在第6行

這是密碼

<?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>"; 

請幫我解決這個錯誤我是一個初學者。

Answer 1

試試這個（我還添加了另一個參數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：創建連接

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

第2步：運行查詢

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

第3步：取記錄

$result = mysqli_fetch_all($query); 

步驟4：顯示記錄

var_dump($result);