使用SELECT结果做出一个表

我想执行下面的SELECT语句：

SELECT count(*) from table WHERE A=1 and date between 2013-02 and 2013-03 
SELECT count(*) from table WHERE A=1 and date between 2013-03 and 2013-04 
SELECT count(*) from table WHERE A=1 and date between 2013-04 and 2013-05 
.... 
SELECT count(*) from table WHERE B=1 and date between 2013-02 and 2013-03 
SELECT count(*) from table WHERE B=1 and date between 2013-03 and 2013-04 
SELECT count(*) from table WHERE B=1 and date between 2013-04 and 2013-05 
...etc

什么是最快的查询，我可以有结果排列在如下表格中：

date |A=1 |B=1 |C=1 |... 
2013-Feb|count|count|count| 
2013-Mar|count|...

来源

2016-09-28 user6571534

从日期提取月份并按月执行GROUP BY。最后算。 –

这是一个数据透视表，有关解决方案的示例，请参阅http://stackoverflow.com/questions/7674786/mysql-pivot-table – hruske

SELECT DATE_FORMAT(date, '%Y-%m'), 
     SUM(CASE WHEN A=1 THEN 1 END) AS Acount, 
     SUM(CASE WHEN B=1 THEN 1 END) AS Bcount, 
     SUM(CASE WHEN C=1 THEN 1 END) AS Ccount, 
     SUM(CASE WHEN D=1 THEN 1 END) AS Dcount, 
     SUM(CASE WHEN E=1 THEN 1 END) AS Ecount 
FROM table 
GROUP BY DATE_FORMAT(date, '%Y-%m')

如果列中的值只有A通过E是0或1（或可能NULL），那么我们就可以简化为：

SELECT DATE_FORMAT(date, '%Y-%m'), 
     SUM(A) AS Acount, 
     SUM(B) AS Bcount, 
     SUM(C) AS Ccount, 
     SUM(D) AS Dcount, 
     SUM(E) AS Ecount 
FROM table 
GROUP BY DATE_FORMAT(date, '%Y-%m')

来源

2016-09-28 08:53:15

嗨，
你可以有GROUP BY功能

SELECT 
    to_char(date,'YYYY-MM') as date, 
    CASE WHEN A=1 THEN COUNT(A) END as A , 
    CASE WHEN B=1 THEN COUNT(B) END as B , 
    CASE WHEN C=1 THEN COUNT(C) END as C , 
    CASE WHEN D=1 THEN COUNT(D) END as D , 
    CASE WHEN E=1 THEN COUNT(E) END as E 
    from table_name group by date order by date

此查询根据年份和月份对列进行分组，并为您提供结果

来源

2016-09-28 08:55:48

使用SELECT结果做出一个表

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