Question

这可能很简单，但我已经仔细检查了文档，无法弄明白。这是源表：

store  item  date       sales
101   00001  27-Oct-12  1
101   00002  27-Oct-12  0
102   00001  27-Oct-12  2  
102   00002  27-Oct-12  1
101   00001  3-Nov-12   0 
101   00002  3-Nov-12   0
102   00001  3-Nov-12   1
102   00002  3-Nov-12   1

...而且我正在尝试编写一个查询，最终得到一份月度销售表，商店有自己的列，例如，只有项目＃00001 -

monthyear  store101sales otherstoresales
10/2012    1             2
11/2012    0             1

这是我试过的代码 -

select distinct 
    year(date) as SalesYear,
    month(date) as SalesMonth,
    sum(case when store in (101) then sales.count else 0 end) as store101sales,
    sum(case when store in (102,103) then sales.count else 0 end) as otherstoresales,
from source_table
where item=00001;
GROUP BY SalesYear, SalesMonth, store101sales, otherstoresales,
ORDER BY SalesYear, SalesMonth, store101sales, otherstoresales;

任何帮助表示感谢;谢谢！

Answer 1

你很接近，不需要GROUP BY聚合，也不确定MySQL是否允许在GROUP BY中使用别名：

select year(date) as SalesYear,
    month(date) as SalesMonth,
    sum(case when store in (101) then sales else 0 end) as store101sales,
    sum(case when store in (102,103) then sales else 0 end) as otherstoresales
from source_table
where item=00001;
GROUP BY year(date), month(date)
ORDER BY SalesYear, SalesMonth, store101sales, otherstoresales;

选择在其自己的列中排列条件的语句

1 个答案: