Question

我是php的新手并尝试制作登录表单，但是在建立连接时它们会出错。我没有得到它。

<?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());
 }

如果有人知道帮助我，请提前。谢谢。

Answer 1

需要将第一个参数作为连接传递给您的数据库名称

mysqli_select_db ($conn, DBNAME );

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

将连接错误检查为

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

Answer 2

试试以下代码：

选择数据库时，您必须先传递$conn，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());
 }

希望这有帮助!!

警告：mysqli_select_db（）期望参数1为mysqli

2 个答案: