如何使用 mysql 填充每组的空记录？

Question

我正在尝试处理 sql 查询以获得我想要的内容。

下面是 table 的架构。

CREATE TABLE MY_LOG (
    RANKING      VARCHAR(20)   
  , DAYOFWEEK      VARCHAR(10) 
  , MENU     VARCHAR(10)   
)

我插入了一些值，如下所示。

Ranking   DAYOFWEEK   MENU
1         MONDAY      PIZZA 
2         MONDAY      ICE CREAM
3         MONDAY      CHICKEN
4         MONDAY      RICE
5         MONDAY      BREAD
1         TUESDAY      PIZZA 
2         TUESDAY      ICE CREAM
3         TUESDAY      CHICKEN
4         TUESDAY      RICE
1         WEDNESDAY      PIZZA 
2         WEDNESDAY      ICE CREAM
3         WEDNESDAY      CHICKEN

如您所见，对于一周中的每一天，排名都与其菜单一起显示。但是，对于星期二和星期三，他们只有四个和三个记录。所以我想插入如下所示的空白记录。

Ranking   DAYOFWEEK   MENU
1         MONDAY      PIZZA 
2         MONDAY      ICE CREAM
3         MONDAY      CHICKEN
4         MONDAY      RICE
5         MONDAY      BREAD
1         TUESDAY      PIZZA 
2         TUESDAY      ICE CREAM
3         TUESDAY      CHICKEN
4         TUESDAY      RICE
5         -            -
1         WEDNESDAY      PIZZA 
2         WEDNESDAY      ICE CREAM
3         WEDNESDAY      CHICKEN
4         -              -
5         -              -

我试图解决这个问题但失败了。如何实现？

Answer 1

您可以尝试将 OUTER JOIN 与 CORSS JOIN 得到结果 RANKING & DAYOFWEEK 笛卡尔积

的子查询一起使用

查询#1

SELECT t1.RANKING,
       t2.DAYOFWEEK,
       t2.MENU
FROM (
  SELECT DISTINCT t1.DAYOFWEEK,t2.RANKING
  FROM MY_LOG t1
  CROSS JOIN MY_LOG t2
) t1 LEFT JOIN  MY_LOG t2
ON t1.RANKING = t2.RANKING 
AND t1.DAYOFWEEK = t2.DAYOFWEEK
ORDER BY t1.DAYOFWEEK,t1.RANKING;

RANKING	DAYOFWEEK	MENU
1	MONDAY	PIZZA
2	MONDAY	ICE CREAM
3	MONDAY	CHICKEN
4	MONDAY	RICE
5	MONDAY	BREAD
1	TUESDAY	PIZZA
2	TUESDAY	ICE CREAM
3	TUESDAY	CHICKEN
4	TUESDAY	RICE
5
1	WEDNESDAY	PIZZA
2	WEDNESDAY	ICE CREAM
3	WEDNESDAY	CHICKEN
4
5

View on DB Fiddle

Answer 2

如果您还想包括一周中的其他日子：

with recursive weekdays as (
   select dayname(curdate()) dn, curdate() as d 
   
   union all 
   
   select dayname(d+1),d+1 
   from weekdays 
   where d+1<date_add(curdate(), interval 7 DAY)
),
oneTofive as (
   select 1 as n union select 2 union select 3 union select 4 union select 5)
select 
   COALESCE(MY_LOG.RANKING, oneTofive.n) as Ranking,
   weekdays.dn as Weekday,
   MY_LOG.Menu
   
from weekdays
cross join oneTofive 
left join MY_LOG ON MY_LOG.DAYOFWEEK = weekdays.dn AND oneTofive.n=MY_LOG.RANKING
order by 
   weekdays.d, 
   COALESCE(MY_LOG.RANKING, oneTofive.n);

输出：

Ranking	Weekday	Menu
1	Sunday
2	Sunday
3	Sunday
4	Sunday
5	Sunday
1	Monday	PIZZA
2	Monday	ICE CREAM
3	Monday	CHICKEN
4	Monday	RICE
5	Monday	BREAD
1	Tuesday	PIZZA
2	Tuesday	ICE CREAM
3	Tuesday	CHICKEN
4	Tuesday	RICE
5	Tuesday
1	Wednesday	PIZZA
2	Wednesday	ICE CREAM
3	Wednesday	CHICKEN
4	Wednesday
5	Wednesday
1	Thursday
2	Thursday
..	etc..

如何使用 mysql 填充每组的空记录？

How to fill the empty record per group using mysql?

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