Question

我有两个表Mention和Retweet

提

ID     USER_SCREEN_NAME     USER_ID    MENTIONED_USER_ID  MENTIONED_USER_NAME
15           user1             1         9                    user 9
16           user 1            1         60                   user 6
18           user 3            3         60                   user 6
17           user 4            4         60                   user 6

转推

ID  USER_ID  USER_SCREEN_NAME RETWEETED_STATUS_ID RETWEETED_USER_STATUS_ID  RETWEETED_USER_NAME
19     3       user 3                  400                    40          user 4
11     1       user 1                  38                     60          user 6
15     2       user 2                  500                    50          user 5
13     4       user 4                  38                     60          user 6

我想要的是一个单独的表，其中包含用户提到或转发的信息，或者两者都包含在表中，因此输出应该是这样的（User_id（M）被提到用户ID和User_ID（R）被转发的用户ID）

 USER_ID (M)  USER_ID (R)   USER_S_NAME    MENTIONED_USER   RETWEETED_USER
    1             1           user 1          user 6            user 6
    1             1           user 1          user 9            Null
    3             3           user 3          user 6            Null
    3             3           user 3           Null             user 4
    4             4           user 4          user 6            user 6
    Null          2           user 2            Null            user 5

但是我得到了这个结果

 USER_ID (M)  USER_ID (R)   USER_S_NAME    MENTIONED_USER   RETWEETED_USER
    1             1           user 1          user 6            user 6
    1             1           user 1          user 9            Null
    3             3           user 3          user 6            Null
    4             4           user 4          user 6            user 6
    Null          2            Null            Null              Null

它适用于提及表，但不适用于转发表这是我的查询

SELECT
    MT.USER_ID AS USER_ID,
    r.USER_ID AS USER__ID,
   mt.USER_SCREEN_NAME AS USER_S_NAME,
    mt.MENTIONED_USER_NAME AS MENTIONED_USER,
    (
        SELECT
            RETWEET.RETWEETED_USER_NAME
        FROM
            RETWEET
        WHERE
            RETWEETED_USER_STATUS_ID = mt.MENTIONED_USER_ID
        AND ROWNUM = 1
        AND USER_ID = MT.USER_ID
    ) AS RETWEETED_USER

FROM
    MENTIONS mt
FULL OUTER JOIN RETWEET r ON MT.USER_ID = r.USER_ID --where MT.M_USER_ID=r.O_USER_ID
GROUP BY
    MT.USER_ID,
  mt.USER_SCREEN_NAME,
    r.USER_ID,
    mt.MENTIONED_USER_NAME,
    MT.MENTIONED_USER_ID,
  r.RETWEETED_USER_STATUS_ID
ORDER BY
    mt.USER_ID

Answer 1

只需添加JOIN子句即可提及/转发ID字段。不需要GROUP BY并且不清楚为什么需要子查询。但是，使用FULL OUTER JOIN无法获得完整的所需结果，因为 nulls 将出现在相应的不匹配的左表或右表中：

SELECT
    mt.USER_ID AS "USER_ID (M)",
    r.USER_ID AS "USER__ID (R)",
    mt.USER_SCREEN_NAME AS "USER_S_NAME",
    mt.MENTIONED_USER_NAME AS "MENTIONED_USER",
    r.RETWEETED_USER_NAME AS "RETWEETED_USER"

FROM
    MENTION mt
FULL OUTER JOIN RETWEET r ON mt.USER_ID = r.USER_ID
   AND mt.MENTIONED_USER_ID = r.RETWEETED_USER_STATUS_ID

ORDER BY mt.user_id, r.user_id

-- USER_ID (M)  USER__ID (R)    USER_S_NAME     MENTIONED_USER  RETWEETED_USER
-- 1            1               user 1          user 6          user 6
-- 1            (null)          user1           user 9          (null)
-- 3            (null)          user 3          user 6          (null)
-- 4            4               user 4          user 6          user 6
-- (null)       2               (null)          (null)          user 5
-- (null)       3               (null)          (null)          user 4

请参阅使用 PostgresSQL 9.3 作为方言的SQL Fiddle，因为已知Oracle不会在小提琴中工作。

在oracle

1 个答案: