我有一张桌子:
NACHRICHT_ID | VERFASSER_USERNAME | BETREFF | TEXT | DATUM | EMPAENGER_ID
------------------------------------------------------------------------------
| | | | 2009-07-01|
1 | h | hfgh | hfgh | 23:15:10 | 31
------------------------------------------------------------------------------
| | | | 2009-07-02|
3 | h | hfgh | hfgh | 00:30:10 | 31
------------------------------------------------------------------------------
| | | | 2009-07-02|
4 | q | hfgh | hi | 11:42:06 | 31
------------------------------------------------------------------------------
| | | | 2009-07-02|
5 | h | hfgh | hfgh | 12:53:56 | 31
------------------------------------------------------------------------------
| | | | 2009-07-02|
6 | h | hfgh | hfgh | 13:19:53 | 31
------------------------------------------------------------------------------
| | | | 2009-07-02|
7 | h | hfgh | hfgh | 13:20:18 | 31
------------------------------------------------------------------------------
| | | | 2009-07-02|
8 | mkd | hfgh | hfgh | 13:47:32 | 31
------------------------------------------------------------------------------
| | | | 2009-07-02|
9 | h | hfgh | hfgh | 16:42:10 | 31
------------------------------------------------------------------------------
我正在尝试删除行NACHRICHT_ID = 3:
DELETE FROM NACHRICHTEN JOIN
NACHRICHTEN_EMPFAENGER_MAP USING (NACHRICHT_ID)
WHERE EMPFAENGER_ID = 31 && NACHRICHT_ID = 3 ;
Run Code Online (Sandbox Code Playgroud)
但它没有用.我究竟做错了什么?
你的问题不清楚.
你有3个表:
要删除新闻吗?然后这样做:
delete from NACHRICHTEN where NACHRICHT_ID = 3;
Run Code Online (Sandbox Code Playgroud)
如果要删除ID为3的新闻项和ID为31的收件人之间的映射,请执行以下操作:
delete from NACHRICHTEN_EMPFAENGER_MAP where NACHRICHT_ID=3 and EMPFAENGER_ID=31;
Run Code Online (Sandbox Code Playgroud)
如果要删除ID为31的实际收件人,请执行此操作(假设该表名为EMPFAENGER)
delete from EMPFAENGER where EMPFAENGER_ID = 31;
Run Code Online (Sandbox Code Playgroud)
如果您想做其他事情,请提出更详细的问题.
| 归档时间: |
|
| 查看次数: |
373 次 |
| 最近记录: |