本文介绍了mysqli_select_db()期望参数1在第8行是mysqli的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我不知道我该怎么办如果有人出了什么问题,请更正并给予解决预先感谢
I Don't know what actually I should do and if there is something wrong somebody please correct it and give itthanks in advance
<?php
$db_host = 'localhost';
$db_user = 'root';
$db_pass = '';
$db_name = 'cms';
$conn = mysqli_connect($db_host,$db_user,$db_pass) or die(mysql_error());
mysqli_select_db($db_name,$conn);
?>
推荐答案
使用此:
$con = mysqli_connect($hostname,$user,$password,$database);
您不需要:
mysqli_select_db
您的代码将是:
$db_host = 'localhost';
$db_user = 'root';
$db_pass = '';
$db_name = 'cms';
$con = mysqli_connect($db_host,$db_user,$db_pass,$db_name);
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