PHP 和 MYSQLI 错误，调用非对象上的成员函数 query()
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PHP and MYSQLI Error, Call to a member function query() on a non-object
提问by dkeeper09
I'm getting the Call to a member function query() on a non-object when I try to call my function.
当我尝试调用我的函数时，我正在调用非对象上的成员函数 query()。
My code looks like this:
我的代码如下所示：
function add_profile() {
$hostname = "localhost";
$dbusername = "username";
$dbname = "dbname";
$dbpassword = "password";
$link = mysqli_connect($hostname, $dbusername, $dbpassword, $dbname);
if (!$link) {
die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error());
}
$sql = "INSERT INTO payment_profiles(id, client_id) VALUES ( '','$profile_id')";
$result = $mysqli->query($sql);
if (!result)
{
echo 'Error: ', $mysqli->error;
}
}
add_profile();
It says my error is on the line: $result = $mysqli->query($sql);
它说我的错误就行了： $result = $mysqli->query($sql);
I'm assuming I'm not calling something properly. Thanks in advance for any help
我假设我没有正确调用某些东西。在此先感谢您的帮助
回答by Saturnix
In your code you're mixing both procedural and object-oriented code. Choose either one or the other. Here's how you would solve the problem the procedural way.
在您的代码中，您混合了面向过程和面向对象的代码。选择其中之一。以下是您如何以程序方式解决问题。
$result = mysqli_query($link, $sql, MYSQLI_USE_RESULT)
I'm getting the Call to a member function query() on a non-object when I try to call my function.
当我尝试调用我的函数时，我正在调用非对象上的成员函数 query()。
That's because the $mysqliobject is not declared anywhere (or is it)? Before you can use $mysqliyou should first create an instance of mysqliand assign it to your object.
那是因为$mysqli对象没有在任何地方声明（或者是）？在您可以使用之前，$mysqli您应该首先创建一个实例mysqli并将其分配给您的对象。
$mysqli = new mysqli("localhost", "my_user", "my_password", "database");
Only then you may call the methods of the mysqli class like $mysqli->query();
只有这样你才能调用 mysqli 类的方法，如 $mysqli->query();
The error you made depends probably on two misconceptions:
您犯的错误可能取决于两个误解：
1) you pasted half of your code from the procedural-style part of the mysqli manual and half from the oop part
1）你从mysqli手册的程序风格部分粘贴了一半的代码，从oop部分粘贴了一半
2) you assume $mysqliis a global object instantiated with $mysqli_connect();. It is not. You should invoke the constructor with the newkeyword if you'd like to use it as an object.
2）您假设$mysqli是一个使用$mysqli_connect();. 它不是。new如果您想将其用作对象，则应使用关键字调用构造函数。
回答by CBroe
$link = mysqli_connect($hostname, $dbusername, $dbpassword, $dbname);
/////
$result = $mysqli->query($sql);
///////
