Question

我有一个名为Leaderboard的表，有2列，PlayerName和PlayerRating。我希望能够访问特定的PlayerName，获得他们的PlayerRating但是他们的排名（例如：他们是第一，第二，第十？）。我做了一些阅读，我确实找到了一种方法，但我似乎无法将其翻译成PHP并实际打印该值。请记住，我对MySQLi或PHP不是很有经验。

我目前正在通过以下方式获取我的PlayerRating：

$CurrentRating = $Connection->query("SELECT CurrentRating FROM Leaderboard WHERE PlayerName='$PlayerName'")->fetch_assoc()["CurrentRating"];

我可以echo $CurrentRating;完成工作。现在找到PlayerRank我发现了几个帖子，解释了我的查询应该如何

SET @rank=0; SELECT @rank:=@rank+1 As rank

所以一开始我想是：

query("SET @rank=0; SELECT @rank:=@rank+1 As rank, CurrentRating FROM Leaderboard [...]

但我似乎无法让它正常工作并获得正确打印的等级。我尝试回复$CurrentRating["rank"]以及以下几种变体：

$CurrentRank = $Connection->query("SET @rank=0; SELECT @rank:=@rank+1 As rank, SELECT CurrentRating FROM Leaderboard WHERE PlayerName='$PlayerName'")->fetch_assoc()["rank"];

有人可以帮我写一下吗？提前谢谢！

Answer 1

当您想要根据CurrentRating进行排名时，您的查询应该是这样的。

<强>查询

    SELECT 
       Leaderboard.PlayerName
     , Leaderboard.PlayerRating
     , (@rank := @rank + 1) AS rank
    FROM 
     Leaderboard
    CROSS JOIN(
      SELECT 
       @rank := 0
    )
      AS
        init_user_variable 
    ORDER BY 
     Leaderboard.CurrentRating DESC

当你想要在哪里使用过滤器排名时，你应该在这样的交付表中使用这个查询。

<强>查询

     SELECT 
       Leaderboard_ranked.PlayerName
     , Leaderboard_ranked.PlayerRating
     , Leaderboard_ranked.rank 
    FROM (

        SELECT 
           Leaderboard.PlayerName
         , Leaderboard.PlayerRating
         , (@rank := @rank + 1) AS rank
        FROM 
         Leaderboard
        CROSS JOIN(
          SELECT 
           @rank := 0
        )
          AS
            init_user_variable 
        ORDER BY 
         Leaderboard.CurrentRating DESC

    ) 
      AS 
       Leaderboard_ranked
    WHERE
     Leaderboard_ranked.PlayerName = 'playername'

PHP MySQLi语法

$connection = mysqli_connect("host/ip", "user", "password", "database");
$result = mysqli_query($connection, [query]);
while($row = mysqli_fetch_array($result)) {
  echo $row['PlayerName'] . " " . $row['PlayerRating'] . " " . $row['rank'] "<br />" 
}

Answer 2

SELECT COUNT(*) FROM Leaderboard WHERE CurrentRating > ?

用户当前评级作为参数应该或多或少

获取特定条目的行号

2 个答案: