警告:mysqli_select_db() 需要 2 个参数,其中 1 个给定 *13*
Warning: mysqli_select_db() expects exactly 2 parameters, 1 given *13*
我正在为我的学校开发一个系统,让教师 post 可以在内联网上查看他们当天收到的任何通知。但是我在测试时遇到了这个错误:
Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\Users\Matthew\Desktop\wamp64\www\my-site\viewguestbook.php on line 23.
页面代码如下:
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="test"; // Database name
$tbl_name="guestbook"; // Table name
// Connect to server and select database.
mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name";
$result=mysqli_query($sql);
while($rows=mysqli_fetch_array($result)){
?>
注意:这与网站上同名的其他问题不同,因为它处于不同的情况。
mysqli_db_select() 方法需要两个参数
连接对象
数据库名称(需要选择)
因此,您应该将连接对象传递给 mysqli_db_select()
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="test"; // Database name
$tbl_name="guestbook"; // Table name
// Connect to server and select database.
$connection = mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db($connection,$db_name)or die("cannot select DB");
$sql="SELECT * FROM $tbl_name";
$result=mysqli_query($connection,$sql);
while($rows=mysqli_fetch_array($result)){
?>
mysqli_select_db()
的第一个参数是连接对象。这是语法:
mysqli_select_db(connection,dbname);
将您的代码更改为:
$con = mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db($con,$db_name)or die("cannot select DB");
也将连接对象添加为 mysqli_query
和 mysqli_fetch_array
函数中的第一个参数。请参考此 link 的语法。
改变
mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db("$db_name")or die("cannot select DB");
到
$connection =mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db($connection ,"$db_name")or die("cannot select DB");
因为mysqli_select_db
也需要连接对象。
我正在为我的学校开发一个系统,让教师 post 可以在内联网上查看他们当天收到的任何通知。但是我在测试时遇到了这个错误:
Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\Users\Matthew\Desktop\wamp64\www\my-site\viewguestbook.php on line 23.
页面代码如下:
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="test"; // Database name
$tbl_name="guestbook"; // Table name
// Connect to server and select database.
mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name";
$result=mysqli_query($sql);
while($rows=mysqli_fetch_array($result)){
?>
注意:这与网站上同名的其他问题不同,因为它处于不同的情况。
mysqli_db_select() 方法需要两个参数
连接对象
数据库名称(需要选择)
因此,您应该将连接对象传递给 mysqli_db_select()
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="test"; // Database name
$tbl_name="guestbook"; // Table name
// Connect to server and select database.
$connection = mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db($connection,$db_name)or die("cannot select DB");
$sql="SELECT * FROM $tbl_name";
$result=mysqli_query($connection,$sql);
while($rows=mysqli_fetch_array($result)){
?>
mysqli_select_db()
的第一个参数是连接对象。这是语法:
mysqli_select_db(connection,dbname);
将您的代码更改为:
$con = mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db($con,$db_name)or die("cannot select DB");
也将连接对象添加为 mysqli_query
和 mysqli_fetch_array
函数中的第一个参数。请参考此 link 的语法。
改变
mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db("$db_name")or die("cannot select DB");
到
$connection =mysqli_connect("$host", "$username", "$password")or die("cannot connect server ");
mysqli_select_db($connection ,"$db_name")or die("cannot select DB");
因为mysqli_select_db
也需要连接对象。