Question

我需要使用删除按钮从数据库中删除新闻文章。我写了所有的代码，但它给了我错误注意：未定义的索引：第5行的C：\ XAMPP \ htdocs \ PortalZaVijesti \ php \ delete.php中的acticleID。我不知道有什么问题所以你看看在我的代码？ admin.php的

$dbc = mysqli_connect("localhost", "root", "", "news_portal") or die(mysql_error());
        $query = "SELECT * FROM news_site;";
        $result = mysqli_query($dbc, $query);
        while ($row = mysqli_fetch_assoc($result)) {
            echo "<p>".$row['title']; 
            echo "</p>
                <form action='php/delete.php' method='post' class='deletePHP'>
                    <input type='hidden' name='articleID' value='".$row['id']."'/>
                    <input type='submit' name='delete' value='delete'/>
                </form>;

delete.php

$dbc = mysqli_connect("localhost", "root", "", "news_portal") or die(mysql_error());
    if(isset($_POST["delete"]) && !empty($_POST["articleID"])){
        $id = mysqli_real_escape_string($dbc, $_POST["acticleID"]);
    }

    $delquery = "DELETE FROM news_site WHERE ID='$id'";
    mysqli_query($dbc, $delquery) or die(mysqli_error($dbc));

    if(mysqli_affected_rows($dbc)){
        echo "It was really successful.";
    }

Answer 1

请修改

行

$id = mysqli_real_escape_string($dbc, $_POST["acticleID"]);

到

$id = mysqli_real_escape_string($dbc, $_POST["articleID"]);

您错误输入了articleId

Answer 2

我可以在delete.php中看到输入错误。

在所示代码的第3行，＆＃34; articleID＆＃34;拼写为＆＃34; acticleID＆＃34;。这就是您获得未定义索引的原因。

变化：

$id = mysqli_real_escape_string($dbc, $_POST["acticleID"]);

为：

$id = mysqli_real_escape_string($dbc, $_POST["articleID"]);

PHP使用按钮从SQL中删除数据库

2 个答案: