Question

因此，我一直在尝试建立一个简单的网站，向MySQL上的数据库发送“喜欢”和“不喜欢”的信息。

根据开发人员网络标签，我不确定我的PHP文件是否已损坏，或者是否存在数据库问题，即请求确实连接。（当我单击“不喜欢/喜欢”按钮时，请求会通过）

不确定要尝试什么。

if (isset($_POST['post_liked'])) {
    $host         = "ip";
    $username     = "admin";
    $password     = "";
    $databasename = "likeanddislike";
    $Connect      = mysqli_connect($host, $username, $password);
    mysqli_select_db($Connect, $databasename);
    mysqli_query($Connect, "update rating_system set total_no_of_votes=total_no_of_votes+1,vote_likes=vote_likes+1");
    $select = mysqli_query($Connect, "SELECT * FROM rating_system");
    while ($row = mysqli_fetch_array($select, 1)) {
        $total_no_votes = $row['total_no_of_votes'];
        $liked          = $row['vote_likes'];
        $disliked       = $row['vote_dislike'];
        echo "<p id='total_rating'>Total Ratings ( " . $total_no_votes . " )</p>";
        echo "<p id='total_like'><img src='attack.png'>" . $liked . "</p><div id='like_bar'></div>";
        echo "<p id='total_dislike'><img src='attack.png'>" . $disliked . " </p><div id='dislike_bar'></div>";
        exit();
    }
}
if (isset($_POST['post_dislike'])) {
    $host         = "ip";
    $username     = "admin";
    $password     = "";
    $databasename = "likeanddislike";
    $connect      = mysqli_connect($host, $username, $password);
    mysqli_select_db($connect, $databasename);
    mysqli_query($connect, "update rating_system set total_no_of_votes=total_no_of_votes+1,vote_dislike=vote_dislike+1");
    $select = mysqli_query($connect, "SELECT * FROM rating_system");
    while ($row = mysqli_fetch_array($select, 1)) {
        $total_no_votes = $row['total_no_of_votes'];
        $liked          = $row['vote_likes'];
        $disliked       = $row['vote_dislike'];

        echo "<p id='total_rating'>Total Ratings ( " . $total_no_votes . " )</p>";
        echo "<p id='total_like'><img src='attack.png'>" . $liked . "</p><div id='like_bar'></div>";
        echo "<p id='total_dislike'><img src='attack.png'>" . $disliked . " </p><div id='dislike_bar'></div>";
        exit();
    }
}
?>

对于HTML文件： Code

PHP文件未更新数据库

0 个答案: