MYSQL LEFT JOIN 与 GROUP BY

MYSQL LEFT JOIN with GROUP BY(MYSQL LEFT JOIN 与 GROUP BY)
本文介绍了MYSQL LEFT JOIN 与 GROUP BY的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

:)我有 2 个查询,我需要加入它们,我需要比较员工的工作时间(取决于活动)与公司在同一活动中在定义的时间段内的总工作时间

:) I have 2 queries, and I need to join them, I need to compare the working time of employee depending on activity with total working time of company in the same activity in defined period

第一个查询是:

SELECT u.login,
       a.article, 
       p.p_article, 
       (SUM(p.p_going) + SUM(p.p_leaving) + SUM(p.p_working)) AS tottime
FROM pos p,users u, articles a
WHERE u.login = p.p_login
AND REPLACE( u.login, '.', '_' ) = 'users_name'
AND p.p_datum >= '2013-04-09'
AND p.p_datum <= '2013-04-16'
AND p.p_article = a.id
GROUP BY a.article

我的第二个查询是:

SELECT a.article, 
       p.p_article, 
       (SUM(p.p_going) + SUM(p.p_leaving) + SUM(p.p_working)) AS tottime
FROM pos p, articles a
WHERE p.p_datum >= '2013-04-09'
AND p.p_datum <= '2013-04-16'
AND p.p_article = a.id
GROUP BY a.article

第一个查询返回我按活动分组的 WORKER 的总工作时间,例如:

The first query returns me total working time of WORKER grouped by activites, for example:

u.login    a.article     p.p_article  tottime
Ivan       Teambuilding    1          3,45
Julie      Social_work     2          5,67

第二个查询返回 COMPANY 按活动分组的总工作时间,例如:

The second query returns me total working time of COMPANY grouped by activites, for example:

a.article     p.p_article  tottime
Teambuilding    1         150
Social_work     2         260

我想要这样的东西,所以我可以将每个活动的工人总时间与特定时期内每个活动的公司工作总时间进行比较:

I want to have something like this, so I can compare the total time of worker per activity with total time of company working hours per activity in specific period:

u.login    a.article     p.p_article  tottime(worker)  tottime(company) 
Ivan       Teambuilding    1          3,45              150  
Julie      Social_work     2          5,67              260

如果是 NULL 值,我想使用 LEFT JOIN.我花了 3 个小时寻找解决方案,但我尝试的一切都不起作用,所以任何帮助都将不胜感激.

In case of the NULL values I would like to use LEFT JOIN. I was searching for the solution for 3 hours, and everything I try is not working, so any help would be appreciated.

推荐答案

您可以将 2 个查询连接在一起作为一对子选择.

You can just join the 2 queries together as a pair of subselects.

类似:-

SELECT Sub1.a, Sub1.b, Sub2.c
FROM (SELECT a, b FROM z) Sub1
INNER JOIN (SELECT a, c FROM y) Sub2
ON Sub1.a = Sub2.a

但是,由于您的第一个示例查询似乎没有带回您所说的详细信息(仅带回 3 列),因此无法真正为您提供更多信息.

However can't really give you more as you first example query doesn't seem to bring back the details you say (only brings back 3 columns).

编辑 - 使用更正后的查询

EDIT - With the corrected queries

SELECT Sub1.login AS User_name, Sub1.article AS Activity, Sub1.p_article AS `Activity id`, Sub1.tottime AS `Totaltime(worker)`, Sub2.tottime AS `Totaltime(company)`
FROM (SELECT u.login,a.article, p.p_article, (SUM(p.p_going) + SUM(p.p_leaving) + SUM(p.p_working)) AS tottime
FROM pos p
INNER JOIN users u ON u.login = p.p_login 
INNER JOIN articles a ON p.p_article = a.id
WHERE REPLACE( u.login, '.', '_' ) = 'users_name'
AND p.p_datum >= '2013-04-09'
AND p.p_datum <= '2013-04-16'
GROUP BY a.article) Sub1
INNER JOIN 
(SELECT a.article, p.p_article, (SUM(p.p_going) + SUM(p.p_leaving) + SUM(p.p_working)) AS tottime
FROM pos p
INNER JOIN articles a ON p.p_article = a.id
WHERE p.p_datum >= '2013-04-09'
AND p.p_datum <= '2013-04-16'
GROUP BY a.article) Sub2
ON Sub1.p_article = Sub2.p_article

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