問題描述

我正在嘗試構建一個簡單的自定義 CMS，但出現錯誤:

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

警告:mysqli_query() 期望參數 1 是 MySQLi，在

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

為什么我會收到這個錯誤?我所有的代碼都已經是 MySQLi 并且我使用了兩個參數，而不是一個.

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>";
    }
}

推薦答案

正如評論中提到的，這是一個范圍界定問題.具體來說，$con 不在您的 getPosts 函數范圍內.

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

您應該將連接對象作為依賴項傳入，例如

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

function getPosts(mysqli $con) {
    // etc

如果您的連接失敗或發生錯誤，我也強烈建議您停止執行.這樣的東西就足夠了

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);

這篇關于警告:mysqli_query() 期望參數 1 是 mysqli，在的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！