如何在php和mysql中处理此类代码
我如何处理这个问题: 警告:mysqli_select_db()要求参数1为mysqli,字符串在第33行的C:\xampp\htdocs\Uregister\accessd.php中给出 这是我的密码:如何在php和mysql中处理此类代码,php,html,mysql,Php,Html,Mysql,我如何处理这个问题: 警告:mysqli_select_db()要求参数1为mysqli,字符串在第33行的C:\xampp\htdocs\Uregister\accessd.php中给出 这是我的密码: mysqli_select_db($database_user_informaion, $user_informaion); $query_user_request = "SELECT * FROM user"; $user_request = mysqli_query($query_use
mysqli_select_db($database_user_informaion, $user_informaion);
$query_user_request = "SELECT * FROM user";
$user_request = mysqli_query($query_user_request, $user_informaion) or die(mysqli_error());
$row_user_request = mysqli_fetch_assoc($user_request);
$totalRows_user_request = mysqli_num_rows($user_request);
看看
示例代码`
谢谢你的回复,但我现在也遇到了同样的错误。。。这是我的代码请显示您的代码。mysqli\u select\u db($database\u user\u information,$user\u information)$query_user_request=“从用户
中选择*”$user\u request=mysqli\u query($query\u user\u request,$user\u information)或die(mysqli\u error())$row_user_request=mysqli_fetch_assoc($user_request)$totalRows\u user\u request=mysqli\u num\u rows($user\u request);
<?php
//Connect to database "one"
$db = mysqli_connect("host", "user", "password", "one")
//Now change the database to "two"
mysqli_select_db($db, "two");
//Your code for whatever you want to do with the database