问题： sql语句查询本周每天的数据量，如果没有该天，数据量默认为0；我的sql语句
SELECT
WEEKDAY(osr.created_date) AS weekDay,
IF (count(1) IS NULL, 0, count(1)) AS num
FROM osr_work AS osr
INNER JOIN el_user AS el ON el.id = 141
WHERE
osr.teacher_id = 141 AND
YEARWEEK(date_format(osr.created_date,'%Y-%m-%d')) = YEARWEEK(now())
GROUP BY WEEKDAY(osr.created_date);结果：
0-6:分别代表周一至周日
我希望得到周一至周日所有的数据，因为周一和周五，周六，周日没有数据，就默认num为0；我该如何写，求指导！！！
SELECT
osr_weekday.weekDay,
count(1) AS num
FROM
osr_work osr,osr_weekday
WHERE
WEEKDAY(osr.created_date) = osr_weekday.weekDay AND
osr.teacher_id = 141 AND
YEARWEEK(date_format(osr.created_date,'%Y-%m-%d')) = YEARWEEK(now())
GROUP BY osr_weekday.weekDay
union ALL
SELECT
osr_weekday.weekDay ,
0
FROM
osr_weekday
WHERE
osr_weekday.weekDay not in (
SELECT
osr_weekday.weekDay
FROM
osr_work osr,
osr_weekday
WHERE
WEEKDAY(osr.created_date) = osr_weekday.weekDay AND
osr.teacher_id = 141 AND
YEARWEEK(date_format(osr.created_date,'%Y-%m-%d')) = YEARWEEK(now())
GROUP BY osr_weekday.weekDay
)
ORDER BY weekDay想把not in优化，折腾了半天还是老样子～有带飞的吗
解决方案2:日期left join 你要查询的条件，然后在从这个获取的数据集中查询数据，然后判断num是不是null就行了
解决方案3:建议建一张辅佐表 weekDay 0-6，num 全部为0， left join辅佐表就可以了。
改用LEFT JOIN就行了呀。