我發現左外加入UNPIVOT結果到INFORMATION_SCHEMA提供的全部字段列表,以便在某些情況下實際解決此問題。
-- test data
CREATE TABLE _t1(name varchar(20),object_id varchar(20),principal_id varchar(20),schema_id varchar(20),parent_object_id varchar(20),type varchar(20),type_desc varchar(20),create_date varchar(20),modify_date varchar(20),is_ms_shipped varchar(20),is_published varchar(20),is_schema_published varchar(20))
INSERT INTO _t1 SELECT 'blah1', 3, NULL, 4, 0, 'blah2', 'blah3', '20100402 16:59:23.267', NULL, 1, 0, 0
-- example
select c.COLUMN_NAME, Value
from INFORMATION_SCHEMA.COLUMNS c
left join (
select * from _t1
) q1
unpivot (Value for COLUMN_NAME in (name,object_id,principal_id,schema_id,parent_object_id,type,type_desc,create_date,modify_date,is_ms_shipped,is_published,is_schema_published)
) t on t.COLUMN_NAME = c.COLUMN_NAME
where c.TABLE_NAME = '_t1'
</pre>
輸出如下:
+----------------------+-----------------------+
| COLUMN_NAME | Value |
+----------------------+-----------------------+
| name | blah1 |
| object_id | 3 |
| principal_id | NULL | <======
| schema_id | 4 |
| parent_object_id | 0 |
| type | blah2 |
| type_desc | blah3 |
| create_date | 20100402 16:59:23.26 |
| modify_date | NULL | <======
| is_ms_shipped | 1 |
| is_published | 0 |
| is_schema_published | 0 |
+----------------------+-----------------------+
CROSS JOIN ... CASE將保留空值。見下面的例子。 – 2009-06-17 13:47:55