警告：mysqli_select_db（）期望参数1是mysqli

Question

我是新来的PHP，并尝试做登录表单，但他们会错误，而建立connection.i没有得到它。

<?php 

// this will avoid mysql_connect() deprecation error. 
error_reporting(~E_DEPRECATED & ~E_NOTICE); 
// but I strongly suggest you to use PDO or MySQLi. 

define('DBHOST', 'localhost'); 
define('DBUSER', 'root'); 
define('DBPASS', ''); 
define('DBNAME', 'simple_login'); 

$conn = mysqli_connect(DBHOST,DBUSER,DBPASS); 
$dbcon = mysqli_select_db(DBNAME,$conn); 

if (!$conn) { 
die("Connection failed : " . mysqli_error()); 
} 

if (!$dbcon) { 
die("Database Connection failed : " . mysqli_error()); 
} 

如果有人知道帮我，请提前。

Answer 1

尝试下面这段代码：

你有当se时通过第一个$conn选中数据库，您已更改参数顺序。

<?php 

// this will avoid mysql_connect() deprecation error. 
error_reporting(~E_DEPRECATED & ~E_NOTICE); 
// but I strongly suggest you to use PDO or MySQLi. 

define('DBHOST', 'localhost'); 
define('DBUSER', 'root'); 
define('DBPASS', ''); 
define('DBNAME', 'simple_login'); 

$conn = mysqli_connect(DBHOST,DBUSER,DBPASS); 
$dbcon = mysqli_select_db($conn,DBNAME); 

if (!$conn) { 
die("Connection failed : " . mysqli_connect_errno()); 
} 

if (!$dbcon) { 
die("Database Connection failed : " . mysqli_connect_errno()); 
} 

希望这个帮助！

Answer 2

需要传递的第一个参数为连接，然后你的数据库名称

mysqli_select_db ($conn, DBNAME); 

读http://php.net/manual/en/mysqli.select-db.php

检查连接误差

/* check connection */ 
    if (mysqli_connect_errno()) { 
      printf("Connect failed: %s\n", mysqli_connect_error()); 
      exit(); 
     }