我是sql的初学者,我不知道如何找到问题的解决方案。
我有一个包含项目的表,一列包含xml值。我想检查一个字段是否存在并更新它。
我找到了修改现有xml值的方法,但是当它为NULL时它不起作用
有我的代码:
SELECT @xml=IDA_Xml,
@isData = IDA_Xml.exist('(/Root/Data[1]/text())[1]')
FROM DATA
WHERE IDA_UId=@guid
If (@xml is NULL)
BEGIN
-- Create a xml value with <Root></Root>
-- It will be update later
-- set isData to 0
END
IF (@isData = 0)
UPDATE ItemData SET IDA_Xml.modify
('
insert <Data Name = "Info">{sql:variable("@number")}</Data>
into (/Root/*)[1]
')
WHERE IDA_UId = @guid;
ELSE
UPDATE ItemData SET IDA_Xml.modify
('
replace value of (/Root/Data[1]/text())[1]
with sql:variable("@number")
')
WHERE IDA_UId = @guid;
答案 0 :(得分:1)
遵循您的代码结构
SELECT @xml=IDA_Xml,
@isData = IDA_Xml.exist('(/Root/Data[1]/text())[1]')
FROM DATA
WHERE IDA_UId=@guid
If (@xml is NULL)
BEGIN
UPDATE ItemData SET IDA_Xml = CAST('<root/>' as XML)
WHERE IDA_UId = @guid;
SET @isData = 0;
END
IF (@isData = 0)
UPDATE ItemData SET IDA_Xml.modify
('
insert <Data Name = "Info">{sql:variable("@number")}</Data>
into (/Root/*)[1]
')
WHERE IDA_UId = @guid;
ELSE
UPDATE ItemData SET IDA_Xml.modify
('
replace value of (/Root/Data[1]/text())[1]
with sql:variable("@number")
')
WHERE IDA_UId = @guid;
但我最好使用单一更新重新构建它,而不是前两种更新
SET IDA_Xml = CAST('construct your entire xml here' as XML)
答案 1 :(得分:1)
在这种情况下,大多数情况下最简单的方法是丢弃现有元素并将其作为新元素插入:
DECLARE @mockup TABLE(ID INT IDENTITY, Descr VARCHAR(100),YourXml XML);
INSERT INTO @mockup VALUES
('element exists','<root><element>test</element><SomeOther>blah</SomeOther></root>')
,('element doesn''t exist','<root><SomeOther>blah</SomeOther></root>')
,('element exists, but empty','<root><element/><SomeOther>blah</SomeOther></root>');
--At the end, all should be `<element>new</element>`
UPDATE @mockup SET YourXml.modify('delete (/root/element)[1]');
UPDATE @mockup SET YourXml.modify('insert <element>new</element> as first into (/root)[1]');
SELECT * FROM @mockup;