mysqli语句出错

Question

I got this error in mysqli

mysqli_select_db() expects exactly 2 parameters, 1 given in /home/u751513549/public_html/recent.php on line 6

This is the code

<?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>";

Please help me to fix this error I am a beginner.

Answer 1

Try this (I have added another parameter to 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));