以最大聚合值顯示查詢結果

Question

我正在嘗試構建一個查詢，以顯示聚合列「NumberOfArticles」中具有最高值的行。但是，由於有多行具有相同的最高值，因此我不確定如何正確限制查詢生成的行數。

SELECT WRT_ID, COUNT(ART_ID) AS NumberOfArticles 
FROM Articles 
GROUP BY WRT_ID 
ORDER BY NumberOfArticles DESC 

Answer 1

SELECT TOP 1 WRT_ID, COUNT(ART_ID) AS NumberOfArticles 
FROM Articles 
GROUP BY WRT_ID 
ORDER BY 2 DESC 

如果有多個行具有最高的NumberOfArticles計數，則會包含關係。如果您不想包含關係，但只有其中一個具有最高NumberOfArticles計數的關係，請將GROUP BY放入子查詢中，並根據需要更改ORDER BY。

SELECT TOP 1 sub.WRT_ID, sub.NumberOfArticles 
FROM (
    SELECT WRT_ID, COUNT(ART_ID) AS NumberOfArticles 
    FROM Articles 
    GROUP BY WRT_ID 
    ) AS sub 
ORDER BY 2 DESC, 1 DESC; 

Answer 2

我想你可能想：

SELECT A.WRT_ID, (SELECT MAX(B.ART_ID) FROM Articles As B WHERE B.WRT_ID = A.WRT_ID) As NumberOfArticles 
FROM Articles A 
ORDER BY 2 Desc 

它使用一個子選擇，以獲得最高Art_Id每個Wrt_Id

我看你有你的樣品，雖然這個問題在使用COUNT提示更多最大值，但如果它是計數，那麼你可以簡單地修改爲：

SELECT A.WRT_ID, (SELECT COUNT(B.ART_ID) FROM Articles As B WHERE B.WRT_ID = A.WRT_ID) As NumberOfArticles 
FROM Articles A 
ORDER BY 2 Desc 

Answer 3

我想這可能是夠你

SELECT WRT_ID, max(ART_ID) AS NumberOfArticles 
FROM Articles 
GROUP BY NumberOfArticles 
ORDER BY NumberOfArticles DESC 

Answer 4

SELECT WRT_ID, COUNT(ART_ID) AS NumberOfArticles 
FROM Articles 
GROUP BY WRT_ID 
HAVING COUNT(ART_ID)=(SELECT MAX(t.c) FROM (SELECT WRT_ID,COUNT(ART_ID) AS c 
              FROM Articles GROUP BY WRT_ID) t) 
ORDER BY NumberOfArticles DESC 

這給你所有的最大行數。