Question

我对mySQL相当新，但真的很绝望。请尽量放轻松。

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

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

它显示我有18个旅程，1晚，6个2晚等等。我希望做出选择，所以将结果结合1至3晚和4至7晚。请帮帮我！

我想要的输出：

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

提前谢谢!! ：）

Answer 1

您可以在GROUP BY子句中使用整数除法：

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

如果您想要只有两个细分，一个用于<=3，另一个用于> 3，那么您可以使用：

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

修改

要获取诸如＆＃39; 1-3＆＃39;，＆＃39; 4-6＆＃39;等桶标记，您可以使用以下查询：

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

SQL：选择特定值输出

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