Question

我编写了以下函数来更新MySQLi数据库表中的值。我没有收到任何错误，但价值也没有更新。我看不出有什么问题。

function update_hangman_highscore($user, $user_highscore){
    echo 'Update highscore called.  High score to update is '.$user_highscore.' for '.$user;
    $db = "localhost";
    $user = "phpuser";
    $pwd = "Ninja1995";
    $database = "ninja_comments";
    $link = mysqli_connect($db, $user, $pwd)or die(mysqli_connect_error());
    mysqli_select_db($link, $database) or die(mysqli_error($link));
   $result = mysqli_query($link, "UPDATE users SET hangman_highscore = '$user_highscore' WHERE username = '$user';") or die(mysqli_error());
}

我正在使用：

调用该函数

if($_SESSION['score'] > $_SESSION['user_highscore']){
    update_hangman_highscore($_SESSION['user'], $_SESSION['score']);
    $_SESSION['message'] = 'Too many wrong guesses.  You died, but you also achieved a new personal highscore!';
}

我在函数中使用了一个echo（参见第一行）来验证函数是否被调用。这也告诉我$ high_score和$ user参数正在正确传递。我也可以用实际值替换这些变量，并且该函数可以正常工作。所以在这一点上，我也没有排除故障。任何帮助将不胜感激。

Answer 1

您正在使用$user变量两次，并且正在重写该值。你应该重命名它。

尝试

function update_hangman_highscore($user, $user_highscore){
    echo 'Update highscore called.  High score to update is '.$user_highscore.' for '.$user;
    $db = "localhost";
    $db_user = "phpuser";
    $pwd = "Ninja1995";
    $database = "ninja_comments";
    $link = mysqli_connect($db, $db_user, $pwd)or die(mysqli_connect_error());
    mysqli_select_db($link, $database) or die(mysqli_error($link));
   $result = mysqli_query($link, "UPDATE users SET hangman_highscore = '$user_highscore' WHERE username = '$user';") or die(mysqli_error());
}

使用PHP更新MySQLi数据库表中的值

1 个答案: