Question

我已经看到很多这样的问题的答案，但没有解释任何问题，因此我无法理解它并用于我的案例。

SELECT st.name, SUM(sa.val), sa.sale_date FROM sales sa
INNER JOIN employee e ON sa.employee_id
INNER JOIN store st ON e.store_id
GROUP BY st.name, sa.sale_date

考虑以下查询结果：

http://sqlfiddle.com/#!9/0faa35/5

+---------+-------------+------------------------+
|  name   | SUM(sa.val) |       sale_date        |
+---------+-------------+------------------------+
| Store 1 |         800 | July, 29 2015 00:00:00 |
| Store 1 |         700 | July, 30 2015 00:00:00 |
| Store 2 |         800 | July, 29 2015 00:00:00 |
| Store 2 |         700 | July, 30 2015 00:00:00 |
+---------+-------------+------------------------+

我需要转置（分组商店名称），使其成为：

+------------+---------+---------+
|    Date    | Store 1 | Store 2 |
+------------+---------+---------+
| 2015-07-29 |     800 |     800 |
| 2015-07-30 |     700 |     700 |
+------------+---------+---------+

Answer 1

这是一个基本的透视查询。在MySQL中，最简单的方法是条件聚合。鉴于SQL Fiddle中的查询，逻辑是：

SELECT sa.sale_date,
       SUM(CASE WHEN st.name = 'Store 1' THEN val ELSE 0 END) as Store1,
       SUM(CASE WHEN st.name = 'Store 2' THEN val ELSE 0 END) as Store2
FROM sales sa INNER JOIN
     employee e
     ON sa.employee_id INNER JOIN
     store st
     ON e.store_id
GROUP BY sa.sale_date;

注意：您应该将实际查询放在您的问题中。

Here是一个SQL小提琴。

Answer 2

您可以使用准备好的查询：

SELECT CONCAT(
  'SELECT sales.sale_date,',
  GROUP_CONCAT('SUM(
     CASE WHEN employee.store_id = ', id , '
     THEN val
     ELSE 0 END) AS `', REPLACE(name, ' ', ''), '`' SEPARATOR ','),
' FROM
  employee
  INNER JOIN sales ON(employee.id = employee_id)
  GROUP BY sales.sale_date'
) INTO @qry FROM (SELECT name,id FROM store) as stores;
PREPARE stmt FROM @qry;
EXECUTE stmt;

小提琴： http://sqlfiddle.com/#!9/0faa35/39/2

转置mysql查询

2 个答案: