Sql XML查询将属性提取为记录/行

我有下面的xml，希望以行的形式获取所有记录。我的xml如下所示

<category ccode="ct8">
  <columns>
    <col colcode="cl_prodn" displaytext="Prodname" responsetype="textbox" tooltip="testts" isrequired="" displayorder="1" />
    <col colcode="cl_descs" displaytext="Descs" responsetype="textarea" tooltip="atser" isrequired="on" displayorder="2" />
  </columns>
</category>

---

分解xml的方法有很多，但我相信这可以帮助您为混合提供更多的概念：

declare @xml xml = 
'<category ccode="ct8">
  <columns>
    <col colcode="cl_prodn" displaytext="Prodname" responsetype="textbox" tooltip="testts" isrequired="" displayorder="1" />
    <col colcode="cl_descs" displaytext="Descs" responsetype="textarea" tooltip="atser" isrequired="on" displayorder="2" />
  </columns>
</category>'

-- get them one at a time by hunting for specific identifier
select @xml.query('(category/columns/col[@displaytext = "Prodname"])')  -- queries for node
select @xml.value('(category/columns/col[@displaytext = "Prodname"]/@colcode)[1]', 'varchar(max)')  -- gives value of node by description
select @xml.value('(category/columns/col[@displaytext = "Descs"]/@colcode)[1]', 'varchar(max)')


-- get them all at once with the (reference).(column).nodes method applied to an xml value in a table.
declare @X table ( x xml);

insert into @X values (@xml)

select 
    t.query('.')
,   t.value('(@colcode)[1]', 'varchar(max)')
from @X a
 cross apply a.x.nodes('//category/columns/col') as n(t)

---

谢谢你的意见。真的很有帮助。

declare @xml xml = 
'<category ccode="ct8">
  <columns>
    <col colcode="cl_prodn" displaytext="Prodname" responsetype="textbox" tooltip="testts" isrequired="" displayorder="1" />
    <col colcode="cl_descs" displaytext="Descs" responsetype="textarea" tooltip="atser" isrequired="on" displayorder="2" />
  </columns>
</category>'

-- get them one at a time by hunting for specific identifier
select @xml.query('(category/columns/col[@displaytext = "Prodname"])')  -- queries for node
select @xml.value('(category/columns/col[@displaytext = "Prodname"]/@colcode)[1]', 'varchar(max)')  -- gives value of node by description
select @xml.value('(category/columns/col[@displaytext = "Descs"]/@colcode)[1]', 'varchar(max)')


-- get them all at once with the (reference).(column).nodes method applied to an xml value in a table.
declare @X table ( x xml);

insert into @X values (@xml)

select 
    t.query('.')
,   t.value('(@colcode)[1]', 'varchar(max)')
from @X a
 cross apply a.x.nodes('//category/columns/col') as n(t)