SQL：选择特定的值输出

Question

我对mySQL相当陌生，但真的很绝望。请尽量让我容易。

我想创建一个关于酒店的报告。现在，我的查询可以告诉我，有多少次预订了多少次不同的旅程。

举个例子：

SELECT i.numberofnights, 
    SUM(CASE WHEN i.nod >='1' THEN '1' ELSE '0' END) as journeyCount 
    FROM itinerary i 
    WHERE i.arrival BETWEEN '2015-01-01' AND '2015-01-31' 
    GROUP BY i.numberofnights 

我的输出：

NoN | journeyCount 
1  | 18 
2  | 6 
3  | 4 
4  | 13 
5  | 12 
6  | 5 
7  | 9 

它表明我有18个行程与1夜，6 2晚等。我的愿望是做出选择，所以结合1至3晚和4至7晚的结果。 请帮帮我！

我想要的输出：

NoN | journeyCount 
1-3 | 28 
4-7 | 39 

预先感谢您！ :)

Answer 1

您可以使用整数除法的GROUP BY子句中：

SELECT (i.numberofnights-1) DIV 3, 
     SUM(CASE WHEN i.nod >=1 THEN 1 ELSE 0 END) as journeyCount 
FROM itinerary i 
WHERE i.arrival BETWEEN '2015-01-01' AND '2015-01-31' 
GROUP BY (i.numberofnights-1) DIV 3 

如果你想只有两个段，一个用于<=3，另一个用于> 3，那么你可以使用：

SELECT IF(CASE WHEN i.numberofnights <= 3 THEN 0 ELSE 1 END = 0, '1-3', '4-7') AS NoN, 
     SUM(CASE WHEN i.nod >=1 THEN 1 ELSE 0 END) as journeyCount 
FROM itinerary i 
WHERE i.arrival BETWEEN '2015-01-01' AND '2015-01-31' 
GROUP BY CASE WHEN i.numberofnights <= 3 THEN 0 ELSE 1 END 

编辑：

为了得到这样的“1-3”桶标签，“4-6”，等你可以使用下面的查询：

SELECT CONCAT(((i.numberofnights-1) DIV 3 + 1) * 3 - 2, 
       '-', 
       ((i.numberofnights-1) DIV 3 + 1) * 3) AS NoN, 
     SUM(CASE WHEN i.nod >=1 THEN 1 ELSE 0 END) as journeyCount 
FROM itinerary i 
WHERE i.arrival BETWEEN '2015-01-01' AND '2015-01-31' 
GROUP BY (i.numberofnights-1) DIV 3 

Demo here