php - 日期作为列(每天创建报告)

Question

我有这个查询生成我想要的结果。我只需要将日期制作成一列

SELECT item, date, SUM(quantity) 
FROM t 
WHERE date between '2015-08-18' and '2015-08-20'
GROUP BY item, date

这是我的 SQL FIDDLE

产生

我想要的结果

谁能至少告诉我如何实现我想要的结果？

Answer 1

像这样使用条件聚合

要从 date 中提取 day 使用 DAY()功能

SELECT item,
       SUM(case when day(date) = 18 then quantity else 0 end) as `18`,
       SUM(case when day(date) = 19 then quantity else 0 end) as `19`,
       SUM(case when day(date) = 20 then quantity else 0 end) as `20`
FROM t 
WHERE date between '2015-08-18' and '2015-08-20'
GROUP BY item

SQL FIDDLE DEMO