查找用戶測試的排名MySQL

-2

這是目前正在獲取所有用戶排名的MySQL查詢。但我想獲取特定的用戶等級。查找用戶測試的排名MySQL

SELECT Percentage_of_Marks,mark,test_id, 
     1+(SELECT count(*) from user_test a 
      WHERE a.Percentage_of_Marks > b.Percentage_of_Marks) as RNK, 
     Percentage_of_Marks 
FROM user_test b 
ORDER BY b.Percentage_of_Marks DESC

我的查詢結果是。

如何找到特定爲test_id的RNK。

來源

2016-11-07 devendra

附加其中爲test_id = '叔044'（from子句和前爲了通過後） – scaisEdge

參見http://meta.stackoverflow.com/questions/333952/why-should-i-provide-an -mcve換什麼，似乎對我將要-A-極簡單的SQL查詢 – Strawberry

SELECT 
    Percentage_of_Marks,mark,test_id, 
    1+ 
    (
     SELECT count(*) 
     from user_test a 
     WHERE a.Percentage_of_Marks > b.Percentage_of_Marks 
    ) as RNK, 
    Percentage_of_Marks 
FROM user_test b 
where b.test_id='$test_id' 
ORDER BY b.Percentage_of_Marks DESC limit 1

來源

2016-11-07 14:00:01 devendra

查找用戶測試的排名MySQL

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