php mysqli_Query 警告：mysqli_query() 期望参数 1 为 mysqli
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Mysqli_Query warning: mysqli_query() expects parameter 1 to be mysqli
提问by shaymaa
I got this error in my code and I don't know how to solve it my code:
我的代码中出现此错误，我不知道如何解决我的代码：
<?php
session_start();
include_once"connect_to_mysql.php";
$db_host = "localhost";
// Place the username for the MySQL database here
$db_username = "root";
// Place the password for the MySQL database here
$db_pass = "****";
// Place the name for the MySQL database here
$db_name = "mrmagicadam";
// Run the actual connection here
$myConnection= mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("mrmagicadam") or die ("no database");
$sqlCommand="SELECT id, linklabel FROM pages ORDER BY pageorder ASC";
$query=mysqli_query($myConnection, $sqlCommand) or die(mysql_error());
$menuDisplay="";
while($row=mysql_fetch_array($query)) {
$pid=$row["id"];
$linklabel=$row["linklabel"];
$menuDisplay='<a href="index.php?pid=' .$pid . '">' .$linklabel. '</a><br/>';
}
mysqli_free_result($query);
?>
and this is error:
这是错误：
Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in C:\xampp\htdocs\limitless\connect_to_mysql.php on line 17
警告：mysqli_query() 期望参数 1 为 mysqli，资源在 C:\xampp\htdocs\limitless\connect_to_mysql.php 第 17 行
What I am doing wrong?
我做错了什么？
回答by F21
You are mixing mysqliand mysqlextensions, which will not work.
You need to use
你需要使用
$myConnection= mysqli_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysqli_select_db($myConnection, "mrmagicadam") or die ("no database");
mysqlihas many improvements over the original mysqlextension, so it is recommended that you use mysqli.
mysqli对原始mysql扩展名有很多改进，所以建议您使用mysqli.
回答by Starx
You are using improper syntax. If you read the docs mysqli_query()you will find that it needs two parameter.
您使用的语法不正确。如果您阅读文档mysqli_query()您会发现它需要两个参数。
mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
混合 mysqli_query ( mysqli $link , 字符串 $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
mysql $linkgenerally means, the resource object of the established mysqli connection to query the database.
mysql $link一般是指，建立mysqli连接的资源对象来查询数据库。
So there are two ways of solving this problem
所以有两种方法可以解决这个问题
mysqli_query();
mysqli_query();
$myConnection= mysqli_connect("$db_host","$db_username","$db_pass", "mrmagicadam") or die ("could not connect to mysql");
$sqlCommand="SELECT id, linklabel FROM pages ORDER BY pageorder ASC";
$query=mysqli_query($myConnection, $sqlCommand) or die(mysqli_error($myConnection));
Or, Using mysql_query()(This is now obselete)
或者，使用mysql_query()（现在已经过时）
$myConnection= mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("mrmagicadam") or die ("no database");
$sqlCommand="SELECT id, linklabel FROM pages ORDER BY pageorder ASC";
$query=mysql_query($sqlCommand) or die(mysql_error());
As pointed out in the comments, be aware of using die to just get the error. It might inadvertently give the viewer some sensitive information .
正如评论中指出的那样，请注意使用 die 来获取错误。它可能会在不经意间给观众一些敏感信息。
