警告:mysqli_query() 期望参数 1 是 mysqli，在 [英] Warning: mysqli_query() expects parameter 1 to be mysqli, null given in

问题描述

我正在尝试构建一个简单的自定义 CMS，但出现错误:

I am trying to build a simple custom CMS, but I'm getting an error:

警告:mysqli_query() 期望参数 1 是 MySQLi，在

Warning: mysqli_query() expects parameter 1 to be MySQLi, null given in

为什么我会收到这个错误?我所有的代码都已经是 MySQLi 并且我使用了两个参数，而不是一个.

Why am I getting this error? All my code is already MySQLi and I am using two parameters, not one.

$con=mysqli_connect("localhost","xxxx","xxxx","xxxxx");

//check connection
if (mysqli_connect_errno($con))
{
    echo "Failed to connect to MySQL:" . mysqli_connect_error();
}

function getPosts() {
    $query = mysqli_query($con,"SELECT * FROM Blog");
    while($row = mysqli_fetch_array($query))
    {
        echo "<div class="blogsnippet">";
        echo "<h4>" . $row['Title'] . "</h4>" . $row['SubHeading'];
        echo "</div>";
    }
}

推荐答案

正如评论中提到的，这是一个范围界定问题.具体来说，$con 不在您的 getPosts 函数范围内.

As mentioned in comments, this is a scoping issue. Specifically, $con is not in scope within your getPosts function.

您应该将连接对象作为依赖项传入，例如

You should pass your connection object in as a dependency, eg

function getPosts(mysqli $con) {
    // etc

如果您的连接失败或发生错误，我也强烈建议您停止执行.这样的东西就足够了

I would also highly recommend halting execution if your connection fails or if errors occur. Something like this should suffice

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); // throw exceptions
$con=mysqli_connect("localhost","xxxx","xxxx","xxxxx");

getPosts($con);

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