我有下表,其中包含XML中的列:
Id | Label | |
---|---|---|
style="文本对齐:右;">1 | Test1 | <terms><destination><email>email11@foo.com</email><email>email12@foo.com</email></destination><content>blabla</content></terms> |
style="text-align: right;">2 | Test2 | > |
select ID, Label,
stuff(
details.query('for $step in /terms/destination/email/text() return concat(", ", string($step))')
.value('.', 'nvarchar(max)'),
1, 2, '')
from @tbl;
请尝试以下解决方案。
由于未提供 DDL 和样本数据填充, 假设">详细信息"列的数据类型为 XML。
.SQL
-- DDL and sample data population, start
DECLARE @tbl TABLE (ID INT IDENTITY PRIMARY KEY, Label VARCHAR(20), Details XML);
INSERT INTO @tbl (Label, Details) VALUES
('Test1',N'<terms><destination><email>email11@foo.com</email><email>email12@foo.com</email></destination><content>blabla</content></terms>'),
('Test2',N'<terms><destination><email>email21@foo.com</email><email>email22@foo.com</email></destination><content>blabla</content></terms>');
-- DDL and sample data population, end
SELECT ID, Label
, REPLACE(Details.query('data(/terms/destination/email/text())').value('.','VARCHAR(MAX)'), SPACE(1), ', ') AS Destination
FROM @tbl;
输出
+----+-------+----------------------------------+
| ID | Label | Destination |
+----+-------+----------------------------------+
| 1 | Test1 | email11@foo.com, email12@foo.com |
| 2 | Test2 | email21@foo.com, email22@foo.com |
+----+-------+----------------------------------+