Question

我的日志表看起来像这样

rpt_id | shipping_id | shop_id | status | create_time
-------------------------------------------------------------
1      | 1           | 600     | 1      | 2013-12-01 01:06:50
2      | 1           | 600     | 0      | 2013-12-01 01:06:55
3      | 1           | 600     | 1      | 2013-12-02 10:00:30
4      | 2           | 600     | 1      | 2013-12-02 10:00:30
5      | 1           | 601     | 1      | 2013-12-02 11:20:10
6      | 2           | 601     | 1      | 2013-12-02 11:20:10
7      | 1           | 601     | 0      | 2013-12-03 09:10:10
8      | 3           | 602     | 1      | 2013-12-03 13:15:58

我想使用单个查询使其看起来像这样

shipping_id | total_activate | total_deactivate
-----------------------------------------------
1           | 2              | 2
2           | 2              | 0
3           | 1              | 0

我该如何查询？

注意：

状态= 1 =激活
状态= 0 =停用
计算总激活/停用规则：查看上面的日志表。 rpt_id 1＆amp; 3，它具有相同的shop_id，shipping_id和status。它应该算作一个。查看结果表。送货ID 1仅由2家商店激活，它们是shop_id 600和601。

你们可以建议我如何进行查询吗？谢谢你的帮助：D

Answer 1

试试这个：

select shipping_id,
       sum(case when status=1 then 1 else 0 end) as total_activate,
       sum(case when status=0 then 1 else 0 end) as total_deactivate
from (select distinct shipping_id, 
                      shop_id, 
                      status
        from test) a
group by shipping_id
order by shipping_id

在小提琴中查看：http://sqlfiddle.com/#!15/f15fd/4

我没有在查询上添加日期，因为它对结果并不重要。

Answer 2

非常感谢...我也已经知道了，你也可以这样做...... thx

SELECT 
    shipping_id, 
    COUNT(DISTINCT CASE WHEN status = 1 THEN shop_id END) AS total_activate, 
    COUNT(DISTINCT CASE WHEN status = 0 THEN shop_id END) AS total_deactivate 
FROM 
    test

GROUP BY 
    shipping_id 
ORDER BY 
    shipping_id

PostgreSQL查询获取数据结果

2 个答案: