如何使用mysql和php获取已排序行的位置

我有一个存储高分的表，以及玩家ID。我希望能够通过球员id提取记录，然后获得排名，或者他们的分数在表中的位置。意思是，基本上我想说“你在第n个”位置，完全基于球员的得分和所有其他得分。例如：如果我在第46位，那么对我来说，位置信息就像你在总分中处于第46位一样。有人能给我举个小例子吗？

当存在重复的列组值时，这将提供重复的列组值：

SELECT t.playerid,
       t.highscore,
       (SELECT COUNT(*)
          FROM TABLE x 
         WHERE x.playerid = t.playerid
           AND x.highscore >= t.highscore) AS rank
  FROM TABLE t
 WHERE t.playerid = ?

IE:如果三名选手在第二名的得分相同，他们的排名值都是2

这将给出一个明显的价值-三名球员并列第二，只有一名将排名第二：

SELECT x.playerid,
       x.highscore,
       x.rank
  FROM (SELECT t.playerid,
               t.highscore,
               @rownum := @rownum + 1 AS rank
          FROM TABLE t
          JOIN (SELECT @rownum := 0) r
      ORDER BY t.highscore DESC) x
 WHERE x.playerid = ?

这里有一个例子

您希望在用户登录时存储其ID，如

 $_SESSION['username'] = $usernamefromdb;
 $_SESSION['id'] = $userid;

然后，您希望在yoru网站上的每个页面上打开一个会话，您将根据$_会话['id']获取动态信息

 session_start();

然后根据用户ID查找数据库中的数据行

 $userid = $_SESSION['id'];
 $rank_query = "SELECT * FROM table_name WHERE id='$userid'";
 $rank_result = mysqli_query($cxn, $rank_query) or die("Couldn't execute query.");
 $row = mysqli_fetch_assoc($rank_result)

然后使用PHP，将第n个位置声明为变量。并从数据库中提取行的总数

$rank = $row['rank'];
$all = $numrows = mysqli_num_rows($result);

呼应出球员的排名

echo $rank . "out of" . $all;

---

有两种方法可以做到这一点：

方法1:

SET @i = 0;

SELECT * FROM 
scores s1 INNER JOIN (SELECT *, @i := @i + 1 AS rank FROM scores ORDER BY score DESC) AS s2 USING (id);

SELECT *, (SELECT COUNT(1) AS num FROM scores WHERE scores.score > s1.score) + 1 AS rank FROM scores AS s1
ORDER BY rank asc

方法2:

SET @i = 0;

SELECT * FROM 
scores s1 INNER JOIN (SELECT *, @i := @i + 1 AS rank FROM scores ORDER BY score DESC) AS s2 USING (id);

SELECT *, (SELECT COUNT(1) AS num FROM scores WHERE scores.score > s1.score) + 1 AS rank FROM scores AS s1
ORDER BY rank asc