所以我有这个代码，我的浏览器给了我警告。代码在这里：
47 mysqli_select_db($database_connHotel, $connHotel);
48 $Result1 = mysqli_query($insertSQL, $connHotel) or die(mysqli_error($connHotel));
警告在这里：
Warning: mysqli_select_db() expects parameter 1 to be mysqli, string given in /Applications/MAMP/htdocs/add_user.php on line 47
Warning: mysqli_query() expects parameter 1 to be mysqli, string given in /Applications/MAMP/htdocs/add_user.php on line 48
我不理解这些警告，并需要帮助我如何在本周末之前为项目修复它。提前感谢您的帮助！
答案 0 :(得分：3)
对于这两个函数来说，MySQL连接对象是第一个变量：
mysqli_select_db($connHotel, $database_connHotel);
另外，
mysqli_query($connHotel, $insertSQL)
<强>参考：
http://php.net/manual/en/mysqli.select-db.php
http://php.net/manual/en/mysqli.query.php
*mysql_* functions are not just converted to mysqli_* functions just by placing an 'i' at the end. Most of them have different set of parameters, please try having a brief search before using mysqli_* functions.
答案 1 :(得分：0)
mysqli_select_db ( mysqli $link , string $dbname )
链接
A link identifier returned by mysqli_connect().
<强> DBNAME
The database name.
所以像这样使用mysqli_select_db($connHotel, $database_connHotel);