PHP脚本错误：mysqli_query（）null给定[关闭]答案

【问题标题】：Php script error: mysqli_query () null given [closed]PHP脚本错误：mysqli_query（）null给定[关闭]
【发布时间】：2022-01-24 22:43:33
【问题描述】：

我为用户注册和表创建编写了一些脚本。但它给了我这个错误。我有点弄清楚是怎么回事，我错误地将 $ 链接传递给我的脚本，但我仍然无法弄清楚如何正确地做到这一点。请帮帮我。

db.php:

require_once './libs/rb-mysql.php';
require_once 'newUserNotification.php';

R::setup('mysql:host=127.0.0.1;dbname=aboutwork', 'root', '');

$link = mysqli_connect('localhost','root','','aboutwork');
if(mysqli_connect_errno())
{
    echo 'Connection to database failed! ('.mysqli_connect_errno().'): '.mysqli_connect_error();
    exit();
}

function register_worker($login, $password)
{

    $user = R::findOne('users', 'login = ?', array($login));
    $id = $user-> id;

    create_new_user_notification($id, $GLOBALS["link"]);
}

newUserNotification.php：

<?php
require_once 'db.php';

$db = $link;

if(isset($_POST['RemoveNotification']))
    echo remove_new_user_notification($_POST['RemoveNotification']);
else
    echo get_new_user_notifications();

function create_new_user_notification($userId)
{
    $sql = "INSERT INTO notifications (user_id) VALUES ('$userId')";
    if(mysqli_query($GLOBALS['db'], $sql)) //error is here
        return 'Successful';
    else
        return "Error: " .mysqli_error($GLOBALS['db']);//and here
}

function remove_new_user_notification($id)
{
    $sql = "DELETE FROM notifications WHERE id='$id'";
    if(mysqli_query($GLOBALS['db'], $sql))
        return "Successful";
    else
        return "Error deleting record: ".mysqli_error($GLOBALS['db']);
}

function get_new_user_notifications()
{
    $sql = "SELECT * FROM notifications";
    $result = mysqli_query($GLOBALS['db'], $sql);

    return json_encode($result->fetch_assoc());
}

错误：

<b>Warning</b>:  mysqli_query() expects parameter 1 to be mysqli, null given in <b>Z:\home\localhost\www\aboutwork\newUserNotification.php</b> on line <b>32</b><br />
<script language=JavaScript src='/denwer/errors/phperror_js.php'></script><!--error--><br />
<b>Fatal error</b>:  Call to a member function fetch_assoc() on a non-object in <b>Z:\home\localhost\www\aboutwork\newUserNotification.php</b> on line <b>34</b><br />
<script language=JavaScript src='/denwer/errors/phperror_js.php'></script>

【问题讨论】：

您像这样启动与数据库的连接 $link = mysqli_connect('localhost','root','','aboutwork');并尝试对 $GLOBALS['db']; 执行查询您应该像这样设置连接 $GLOBALS['db']= mysqli_connect('localhost','root','','aboutwork');第二个错误是关于获取结果的。在获取之前始终检查结果是否存在。像这样： if($result) {$data=json_encode($result->fetch_assoc());} return $data;
您根本不应该使用 $GLOBALS。将 db 连接作为参数传递给函数...
我需要如何将 $GLOBALS['db'] 传递给 newUserNotification.php？我已经按照你说的做了，但我有同样的错误。我只是这样写： function get_new_user_notifications() { $sql = "SELECT * FROM notifications"; $result = mysqli_query($GLOBALS['db'], $sql); if($result) 返回 json_encode($result->fetch_assoc());否则返回“获取通知失败：”.mysqli_error($GLOBALS['db']); }
请在您的问题中更加具体，并且不要使用 cmets 进一步提问

标签： php mysqli

【解决方案1】：

试试这个修改后的代码：

<?php
require_once 'db.php';

$db = $link;

if (isset($_POST['RemoveNotification'])) {
    echo remove_new_user_notification($link, $_POST['RemoveNotification']);
} else {
    echo get_new_user_notifications($link);
}

function create_new_user_notification($link, $userId) {
    $sql = "INSERT INTO notifications (user_id) VALUES ('$userId')";
    if(mysqli_query($link, $sql)) //error is here
        return 'Successful';
    else
        return "Error: " .mysqli_error($link);//and here
}

function remove_new_user_notification($link, $id) {
    $sql = "DELETE FROM notifications WHERE id='$id'";
    if(mysqli_query($link, $sql))
        return "Successful";
    else
        return "Error deleting record: ".mysqli_error($link);
}

function get_new_user_notifications($link) {
    $sql = "SELECT * FROM notifications";
    $result = mysqli_query($link, $sql);

    return json_encode($result->fetch_assoc());
}

【讨论】：

我试过了，没有任何改变