[英]MySql - Sql query or procedure to generate report by transposing rows into columns
[英]Transposing mysql query
对于这样的问题,我已经看到许多答案,但是没有解释任何查询,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 |
+---------+-------------+------------------------+
我需要对它进行转置(对商店名称进行分组),使其变为:
+------------+---------+---------+
| Date | Store 1 | Store 2 |
+------------+---------+---------+
| 2015-07-29 | 800 | 800 |
| 2015-07-30 | 700 | 700 |
+------------+---------+---------+
这是一个基本的数据透视查询。 在MySQL中,最简单的方法是条件聚合。 给定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:您应该在问题中添加实际查询。
这是一个SQL Fiddle。
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
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