这是
XQuery
为了拯救你:
DECLARE @xml XML=
N'<questions>
<question>
<text>Some text</text>
<questionType></questionType>
<questionSubType>ABC</questionSubType>
<options>
<option subType="001">
<text>Y</text>
<mappedCodes>
<code>1</code>
</mappedCodes>
</option>
<option subType="001">
<text>N</text>
<mappedCodes>
<code>2</code>
</mappedCodes>
</option>
<option subType="002">
<text>Y</text>
<mappedCodes>
<code>1</code>
</mappedCodes>
</option>
</options>
</question>
<question>
<text>Some more text</text>
<questionType></questionType>
<questionSubType>DEF</questionSubType>
<options>
<option subType="001">
<text>Single</text>
<mappedCodes>
<code>PL0157</code>
</mappedCodes>
</option>
<option subType="001">
<text>Married</text>
<mappedCodes>
<code>PD0241</code>
</mappedCodes>
</option>
<option subType="002">
<text>Single</text>
<mappedCodes>
<code>PL1157</code>
</mappedCodes>
</option>
<option subType="002">
<text>Married</text>
<mappedCodes>
<code>PD1241</code>
</mappedCodes>
</option>
</options>
</question>
<question>
<text>Some last text</text>
<questionType></questionType>
<questionSubType>ABC</questionSubType>
<options>
<option subType="001">
<text>T</text>
<mappedCodes>
<code>2</code>
</mappedCodes>
</option>
<option subType="002">
<text>V</text>
<mappedCodes>
<code>2</code>
</mappedCodes>
</option>
</options>
</question>
</questions>';
DECLARE @subType varchar(5) = '001'
,@questionSubType varchar(5) = 'ABC';
--那个
将遍历XML并添加具有给定类型的所有问题,然后添加除
<options>
. 最后一个节点将使用筛选器谓词再次添加:
SELECT @xml.query
('<questions>
{
for $q in /questions/question[(questionSubType/text())[1]=sql:variable("@questionSubType")]
return
<question>
{
$q/*[local-name()!="options"]
}
{
$q/options/option[@subType=sql:variable("@subType")]
}
</question>
}
</questions>
');