1

つまり、2つのテーブルAとbがあります。

表Aは

+------+---------------+
|MANUID|LMD            |
+------+---------------+
|1,001 |10/18/2013 7:13|
+------+---------------+
|1,001 |10/18/2013 7:13|
+------+---------------+
|1,001 |10/18/2013 7:13|
+------+---------------+
|1,001 |10/18/2013 7:13|
+------+---------------+
|1,001 |10/18/2013 7:13|
+------+---------------+
|1,001 |10/18/2013 7:13|
+------+---------------+
|1,001 |10/18/2013 7:13|
+------+---------------+

表Bは

+------+---------------------------------+
|MANUID|MANUFACTURERDESC                 |
+------+---------------------------------+
|1,001 |ACTION CONSTRUCTION EQUIPMENT LTD|
+------+---------------------------------+
|1,002 |AJAX FIORI ENGG (INDIA) PVT  LTD |
+------+---------------------------------+
|1,003 |APOLLO CONST. EQUIPMENT  PVT LTD |
+------+---------------------------------+
|1,004 |APOLLO INDUSTRIAL EQUIP.  PVT LTD|
+------+---------------------------------+
|1,006 |ASHOK LEYLAND                    |
+------+---------------------------------+
|1,007 |ATUL AUTO LTD                    |
+------+---------------------------------+
|1,009 |ATLAS COPCO INDIA LTD            |
+------+---------------------------------+

テーブルAでは利用できないテーブルBからMANUIDを削除したいので、上記から、最終的なテーブルBはMANUIDに1,001しかありません

4

4 に答える 4

3 
SELECT * FROM Table_B
WHERE MANUID IN
(SELECT MANUID FROM Table_A)

また

SELECT * FROM Table_B
JOIN Table_A ON Table_B.MANUID = Table_A.MANUID

または、レコードを削除する場合:

DELETE * FROM Table_B
WHERE MANUID NOT IN
(SELECT MANUID FROM Table_A)
于 2013-11-12T13:29:56.410 に答える
0

SQLクエリを使用している場合これを試してください：

select tb.*
from Table B as tb, Table A as ta
where tb.MANUID = ta.MANUID
于 2013-11-12T13:33:09.997 に答える