算作零，如果行不使用GROUP BY与LEFT JOIN存在

Question

我与MySQL练习，停留在一个场景如下：

我有两个表如下。

Table 1 : 
    Distinct_dates having DATE as my_date column name: 

    Table 2:my_records having following fields 

    record_id, 
    record_submitter_name, 
    record_submit_date as DATETIME 

    I need to get the records submitted by each submitter for each unique month and it should return 0 if no record found for a month. 

    For example: the below query gives me distinct months from distinct_dates table. 
    select distinct(DATE_FORMAT(my_date,'%m-%Y')) from distinct_dates; 
    +--------------------------------+ 
    | (DATE_FORMAT(my_date,'%m-%Y')) | 
    +--------------------------------+ 
    | 01-2016      | 
    | 01-2017      | 
    | 02-2017      | 
    | 03-2017      | 
    +--------------------------------+ 
    4 rows in set (0.00 sec) 

I am able to get the count of Records in each month by submitter and not getting Zero if No Record found for the particular month? 

select distinct(DATE_FORMAT(record_submit_date,'%m-%Y')),count(record_id),record_submitter_name from my_records group by DATE_FORMAT(record_submit_date,'%m-%y'),records_submitter_name; 

I have basic understanding of JOINS but not able to get the correct values ,Any help or pointer will be appreciable. 

我需要每个record_submitter_name得到记录的计数为每个不同的一个月，如果没有记录在特定月份提交的用户，应该给予零？

进一步使用LEFT加入我能够获得NULL值为不同日期的数据不存在的月份，下面是我查询。

select distinct(DATE_FORMAT(d.my_date,'%m-%Y')),count(m.record_id), m.record_submitter_name from distinct_dates d left join my_records m on DATE_FORMAT(m.record_submit_date,'%m-%Y') = DATE_FORMAT(d.my_date,'%m-%Y') and record_submitter_name='xyz' group by d.my_date; 

+----------------------------------+----------------+-------------------+ 
| (DATE_FORMAT(d.my_date,'%m-%Y')) | count(m.record_id) | record_submitter_name | 
+----------------------------------+----------------+-------------------+ 
| 01-2016       |    1 | xyz    | 
| 01-2017       |    1 | xyz    | 
| 02-2017       |    0 | NULL    | 
| 03-2017       |    0 | NULL    | 
+----------------------------------+----------------+-------------------+ 

但使用GROUP BY子句时，它不工作： 使用集团通过近，我还没有收到所要的结果：

select distinct(DATE_FORMAT(d.my_date,'%m-%Y')),count(m.record_id), m.record_submitter_name from distinct_dates d left join my_records m on DATE_FORMAT(m.record_submit_date,'%m-%Y') = DATE_FORMAT(d.my_date,'%m-%Y') group by d.my_date,record_submitter_name; 

Following NULL results was not in the output .... 
    . 
     +----------------------------------+----------------+-------------------+ 
     | (DATE_FORMAT(d.my_date,'%m-%Y')) | count(m.record_id) | record_submitter_name | 
     +----------------------------------+----------------+-------------------+ 
     | 01-2016       |    1 | xyz    | 
     | 01-2017       |    1 | xyz    | 
     | 02-2017       |    0 | NULL    | 
     | 03-2017       |    0 | NULL    | 
     +----------------------------------+----------------+-------------------+ 

是不是有什么毛病GROUP BY子句中使用LEFT JOIN ？

感谢

Answer 1

你需要两个主表的日期法师&另一个是员工主，并创建两个表的交叉产品，您将获得每月明智&年全体员工的条目。

这里我有两个表TABLE_A日期主&表-B作为员工主

创建跨产品： SELECT DISTINCT A.MONTH,A.YEAR,A.MONTH_YEAR,B.EMP_ID,B.EMP_NAME FROM TABLE_A A JOIN TABLE_B B ON 1=1

现在执行左连接在桌子上，你需要做的次数该员工的记录。

SELECT DISTINCT A.MONTH,A.YEAR,A.MONTH_YEAR,B.EMP_ID,B.EMP_NAME 
,COUNT(C.ASSESMENT)AS COUNTS 
FROM TABLE_A A 
JOIN TABLE_B B 
ON 1=1 
LEFT JOIN TABLE_C C 
ON B.EMP_ID=C.EMP_ID 
AND A.MONTH_YEAR=C.MONTH_YEAR 
GROUP BY A.MONTH,A.YEAR,A.MONTH_YEAR,B.EMP_ID,B.EMP_NAME 
--ORDER BY EMP_ID,YEAR,MONTH 

其结果将是这样的： - Result set

Answer 2

使用左加入，

你应该创建一个包含所有日期的主表。

说你的记录是1900年，那么这张表应该有1900-01-01到2100-01-01的记录。

然后，

select 
    distinct(DATE_FORMAT(m.record_submit_date,'%m-%Y')),count(m.record_id),m.record_submitter_name 
from 
    dates d 
    left outer join 
    my_records m on DATE_FORMAT(m.record_submit_date,'%m-%Y') = d.date 
group by d.date,m.records_submitter_name; 

Answer 3

我创造了另一个员工的主表由Sunjit的建议，并根据需要下面的查询工作。

SELECT distinct(DATE_FORMAT(A.my_date,'%m-%Y')),B.EMP_NAME ,COUNT(record_id)AS record_count FROM distinct_dates A JOIN employees B ON 1=1 LEFT JOIN my_records C ON B.emp_name=C.record_submitter_name AND DATE_FORMAT(A.my_date,'%m-%Y')=DATE_FORMAT(C.record_submit_d‌​ate,'%m-%Y') GROUP BY A.my_date,'%m-%Y',B.emp_name;