Question

我觉得这应该很简单，但是我在寻找最佳方法方面遇到了问题

此subitem_to_item_status表中有许多具有状态的子项，但是我需要为所有尚未存在的子项插入一条记录。我不需要更新其中的内容，我实际上只是在尝试复制“插入/忽略”

子项目的样本数据

ITEM_SUBITEMT

item_subitem_id   |   creator_identifier
-----------------------------------------
12                      12345
13                      12345
14                      12345
15                      12345
16                      12345
17                      12345
18                      12345
19                      12345
20                      12345
21                      12345
22                      12345

SUBITEM_TO_ITEM_STATUS

SUBITEM_ID   |   ITEM_STATUS_ID   |   CREATED_BY_IDENTIFIER   
------------------------------------------------------------
12                  1                   12345
15                  1                   12345
16                  1                   12345
20                  1                   12345

所以有了这些数据，我只需要插入第一个表中没有ID 12,15,16或20的记录

    MERGE INTO schema.SUBITEM_TO_ITEM_STATUS (SUBITEM_ID,ITEM_STATUS_ID, CREATED_BY_IDENTIFIER,ROWCHANGE,CREATED_AT) AS T
    USING( (SELECT ITEM_SUBITEM_ID, 1, CREATOR_IDENTIFIER, NOW(),NOW() FROM schema.ITEM_SUBITEMT) AS S
    ON S.ITEM_SUBITEM_ID = T.SUBITEM_ID
    WHEN NOT MATCHED THEN INSERT;

Answer 1

错误的MERGE语句。
试试这个：

MERGE INTO schema.SUBITEM_TO_ITEM_STATUS T
USING (SELECT ITEM_SUBITEM_ID, 1 AS ITEM_STATUS_ID, creator_identifier FROM schema.ITEM_SUBITEMT) S ON S.ITEM_SUBITEM_ID = T.SUBITEM_ID
WHEN NOT MATCHED THEN 
INSERT (SUBITEM_ID, ITEM_STATUS_ID, CREATED_BY_IDENTIFIER) 
VALUES (S.item_subitem_id, S.ITEM_STATUS_ID, S.creator_identifier);

Answer 2

为什么不使用insert？

insert into SUBITEM_TO_ITEM_STATUS (SUBITEM_ID, ITEM_STATUS_ID, CREATED_BY_IDENTIFIER)
   select item_subitem_id, 1, creator_identifier
   from ITEM_SUBITEMT
   where item_subitem_id not in (12, 15, 16, 20);

如果您的意思是不想在表中重复，则可以尝试：

insert into SUBITEM_TO_ITEM_STATUS (SUBITEM_ID, ITEM_STATUS_ID, CREATED_BY_IDENTIFIER)
   select item_subitem_id, 1, creator_identifier
   from ITEM_SUBITEMT s
   where not exists (select 1
                     from SUBITEM_TO_ITEM_STATUS tis
                     where tis.item_subitem_id = s.item_subitem_id
                    );

合并或插入对db2的忽略

2 个答案: