我发现左外连接 UNPIVOT 结果到完整的字段列表,方便地从 INFORMATION_SCHEMA 中提取,在某些情况下是解决此问题的实用答案。
-- 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>
输出看起来像:
+----------+---------- +
| COLUMN_NAME |价值 |
+----------------------------------+----------+
|姓名 |废话1 |
| object_id | 3 |
| principal_id |空 |