求一个Mysql语句 查询出当前周的数据按照天分组

Question

SELECT
DATE_FORMAT(uploadTime_beg, "%Y-%m-%d") as time,
SUM(field01) as sumStatus1,
SUM(field02) as sumStatus2,
SUM(field03) as sumStatus3,
SUM(field04) as sumStatus4,
SUM(field05) as sumStatus5
FROM
health_realdata
WHERE DATE_FORMAT(uploadTime_beg,'%Y%u') = DATE_FORMAT(CURDATE(),'%Y%u')
GROUP BY DATE_FORMAT(uploadTime_beg,"%Y-%m-%d");

    这个SQL查出的数据如下：
    ![图片说明](https://img-ask.csdn.net/upload/201511/24/1448359276_377950.png)

    我想要的效果是：虽然数据库中没有2015-11-25的数据，但是也要查出来，只是后面的值显示为0.

    请问在MYSQL中如何实现，上面的SQL如何修改呢。

    我看到网上说需要一个日历辅助表，说的不是很详细，我试了一下好像有问题，
如果这种方法可行，还请大牛们写详细点，谢谢。

Answer 1

思路：
1、将你上边查询出来的结果跟你说的那个有从2015-11-01 到2015-11-30的日历辅助表进行右连接查询
2、将1中查询出的结果进行再次的查询处理，将null值更换为0；

下边的为代码示意，没有进行调试，应该会有问题，大体结构是这样的：

     SELECT o.time, 
    case when o.sumStatus1 is null then 0 else o.sumStatus1 end as sumStatus1,
    case when o.sumStatus2 is null then 0 else o.sumStatus2 end as sumStatus2,
    case when o.sumStatus3 is null then 0 else o.sumStatus3 end as sumStatus3,
    case when o.sumStatus4 is null then 0 else o.sumStatus4 end as sumStatus4,
    case when o.sumStatus5 is null then 0 else o.sumStatus5 end as sumStatus5
    (
        (
        SELECT
        DATE_FORMAT(uploadTime_beg, "%Y-%m-%d") as time,
        SUM(field01) as sumStatus1,
        SUM(field02) as sumStatus2,
        SUM(field03) as sumStatus3,
        SUM(field04) as sumStatus4,
        SUM(field05) as sumStatus5
        FROM
        health_realdata
        WHERE DATE_FORMAT(uploadTime_beg,'%Y%u') = DATE_FORMAT(CURDATE(),'%Y%u')
        GROUP BY DATE_FORMAT(uploadTime_beg,"%Y-%m-%d");
        ) as a 
        right join 表d on a.time = d.time
    ) o