php mysql_select_db() 期望参数 2 是资源，给定的对象
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mysql_select_db() expects parameter 2 to be resource, object given
提问by Merna
I'm new in using PHP. I made something simple to connect to MySQL and select a database:
我是使用 PHP 的新手。我做了一些简单的事情来连接到 MySQL 并选择一个数据库：
$conn = mysqli_connect($db_host, $db_admin, $db_pass) or die(mysql_error());
// these variables are previously declared and initialized
$selected_db = mysql_select_db($db_name, $conn) or die(mysql_error());
When I tested it, I got a successfully-established connection and the following warning:
当我测试它时，我得到了一个成功建立的连接和以下警告：
mysql_select_db() expects parameter 2 to be resource, object given
Why did this happen? How can I fix it?
为什么会这样？我该如何解决？
回答by Fabio
You are using both mysqliand mysqlsimply change
您正在使用两者mysqli，mysql只需更改
mysql_select_db()
mysql_select_db()
With
和
mysqli_select_db
Referencehttp://php.net/manual/en/mysqli.select-db.php
参考http://php.net/manual/en/mysqli.select-db.php
updated
updated
When you use mysql_select_dbyou are supposed to use mysqlapi and so you have to exatibilish connection to database with mysql sintax mysql_connectReference
当你使用mysql_select_db你应该使用mysql的API，所以你必须exatibilish连接到数据库与MySQL sintaxmysql_connect参考
Mysql is now deprecated so it's correct either to use mysqlior PDO
Mysql 现在已被弃用，因此使用mysqli或使用它都是正确的PDO
回答by Chris Forrence
In addition to using mysqli_* consistently (as mentioned in Fabio's answer), there is an additional problem (and a suggestion):
除了始终如一地使用 mysqli_* （如Fabio 的回答中提到的），还有一个额外的问题（和一个建议）：
While the parameter order in
mysql_select_databaseare database name, connection, the order of parameters inmysqli_select_dbare connection, database name.mysqli_select_db($conn, $db_name);As a suggestion,
mysqli_connectincludes an optional fourth parameter to connect to a particular database. This would allow you to avoid callingmysql_select_dbaltogether.$conn = mysqli_connect($db_host, $db_admin, $db_pass, $db_name) or die(mysqli_connect_error());
其中参数顺序
mysql_select_database是数据库名称、连接，参数顺序mysqli_select_db是连接、数据库名称。mysqli_select_db($conn, $db_name);作为建议，
mysqli_connect包括可选的第四个参数以连接到特定数据库。这将使您mysql_select_db完全避免调用。$conn = mysqli_connect($db_host, $db_admin, $db_pass, $db_name) or die(mysqli_connect_error());
回答by pratyay
You have to change mysql_select_dbto mysqli_select_dbas pointed out by Fabiobut you'll get an error
你必须改变mysql_select_db到mysqli_select_db如指出的Fabio，但你会得到一个错误
mysqli_select_db() expects parameter 1 to be mysqli, string given
For someone experiencing this, reverse the order of parameters, like for in this case give
对于遇到这种情况的人，请颠倒参数的顺序，例如在这种情况下给出
$selected_db = mysqli_select_db($conn, $db_name)
