Question

我正试图绕过一个查询而我无法理解。如果有人给我指针，我将不胜感激。作为我想要实现的一个简单示例，我在数据库中有这些记录

Score|Ranking
-------------
100  |0
200  |0
300  |0

我希望排名字段包含1,2,3根据谁获得最高分，所以结果应为：

Score|Ranking
-------------
100  |3
200  |2
300  |1

目前，我正在为所有这些记录做下一个循环，但实际上可能是几千个 - 这可能需要永远！有没有人对魔法查询有所了解，可以一次性做到这一点？

Answer 1

这是一种方法：

SET @r=0;
UPDATE table SET Ranking= @r:= (@r+1) ORDER BY Score DESC;

/* use this if you just want to pull it from the db, but don't update anything */
SET @r=0;
SELECT *, @r:= (@r+1) as Ranking FROM table ORDER BY Score DESC;

Answer 2

在MySQL中，您可以使用row_number 。

Here's an example在SELECT中使用它：

select @rownum:=@rownum+1 ‘rank’, p.* 
from player p, (SELECT @rownum:=0) r 
order by score desc;

如果您INSERT INTO使用SELECT这样的话，您将获得排名。

Answer 3

这会创建一个内联更新语句，按照变量@rc对您的玩家进行排名。我在非常类似的情况下多次使用它，它运行良好并且保留在数据库端。

SET @rc = 0;
UPDATE players JOIN (SELECT @rc := @rc + 1 AS rank, id FROM players ORDER BY rank DESC)
AS order USING(id) SET players.rank = order.rank;

id被假定为players表的主键。

Answer 4

SET @r = 0;
UPDATE players JOIN (SELECT @r := @r + 1 AS rank, id FROM players ORDER BY rank DESC)
AS sorted USING(id) SET players.rank = sorted.rank;

Answer 5

我告诉你我的方式[用于间隔sql更新功能]

选择

set @currentRank = 0,
    @lastRating = null,
    @rowNumber = 1;
select
    *,
    @currentRank := if(@lastRating = `score`, @currentRank, @rowNumber) `rank`,
    @rowNumber := @rowNumber + if(@lastRating = `score`, 0, 1) `rowNumber`,
    @lastRating := `score`
from `table`
order by `score` desc

更新

set @currentRank = 0,
    @lastRating = null,
    @rowNumber = 1;
update 
    `table` r
    inner join (
        select
            `primaryID`,
            @currentRank := if(@lastRating = `score`, @currentRank, @rowNumber) `rank`,
            @rowNumber := @rowNumber + if(@lastRating = `score`, 0, 1) `rowNumber`,
            @lastRating := `score`
        from `table`
        order by `score` desc
    ) var on
        var.`primaryID` = r.`primaryID`
set
    r.`rank` = var.`rank`

我没有对此测试进行任何性能检查，除非测试它是否有效

Answer 6

如果您使用的是MySQL 8，那么可以使用新功能RANK()

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根据您希望以均匀分数显示排名的方式，这样您还可以签出DENSE_RANK()

作为更新：

  String find(List<ProjectCode> projectCodeList, String projectTitle) {
    return projectCodeList
        .firstWhere(
            (projectCode) => projectCode.projectTitle.contains(projectTitle))
        .id;
  }

用于存储排名位置的MySQL更新语句

6 个答案: