函数中的Mysqli查询-PHP [英] Mysqli query in function - PHP

问题描述

我在 functions.php 中有函数列表.我正在从 Mysql 升级到 Mysqli ，因为我刚刚学习的 Mysql 现已贬值.

I have a list of functions in functions.php. I'm upgrading from Mysql to Mysqli because I just learning Mysql is now depreciated.

我在顶层connect.php文件中声明我的连接.需要第一个文件.

无论如何，我所有的函数都使用mysql_query("QUERY")，并且始终可以正常工作.现在，我将它们全部更改为:

Anyway back to the point all my functions use mysql_query("QUERY") and that always worked fine. Now I changed all the them to:

$con->query("QUERY") // ($con is my connection variable)

现在我遇到一个 致命错误:第241行，在C:\ wamp \ www \ PHP \ functions.php中的非对象上调用成员函数query() > .

所以我不明白为什么要查询我是否在整个文件中声明了变量.我不确定它在任何地方都可以访问.这使我的网站处于暂停状态，直到我可以解决此问题为止.这是 functions.php

function getSiteName()
{
    $row = $con->query("SELECT * FROM siteinfo")->fetch_array();
    return $row["siteName"];
}

我的联系:

global $con ;
$con = new mysqli("localhost", "itunes89", "XXXX","projectanvil") or die("Sorry, were having server connection issues. Please try again later.");

谢谢你们！ :D 我的网站出现此错误

Thanks guys! :D My site where I'm having this error

推荐答案

这是范围可变问题.您需要将$conn传递给getSiteName():

That's a variable scope problem. You need to pass $conn to getSiteName():

function getSiteName($con) {
    $row = $con->query("SELECT * FROM siteinfo")->fetch_array();
    return $row["siteName"];
}

$name = getSiteName($con);

或使用带有构造函数注入的类:

Or using a class with constructor injection:

class MyThing
{
    protected $con;

    public function __construct($con) {
        $this->con = $con;
    }

    public function getSiteName() {
        $row = $this->con->query("SELECT * FROM siteinfo")->fetch_array();
        return $row["siteName"];
    }
}

$obj = new MyThing($con);
$name = $obj->getSiteName();

在整个类中，您可以使用$this->con来访问连接.

Throughout the class you can use $this->con to access the connection.

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