我正在尝试使用XQuery对XML文档进行排序。第一级排序是按父元素之一进行排序。
第二级排序是链接的子元素之一。
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<FEATURES xmlns="http://www.example.com/" xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<FEATURE>
<abbreviation>CANAL</abbreviation>
<LINKED_PROPERTIES>
<LP>
<abbreviation>zltinf</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>altalt</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>betbet</abbreviation>
<category>C</category>
<remarks />
</LP>
</LINKED_PROPERTIES>
<description>Water Canal</description>
<code>23</code>
<relation_type />
</FEATURE>
<FEATURE>
<abbreviation>AREA</abbreviation>
<LINKED_PROPERTIES>
<LP>
<abbreviation>zltzlt</abbreviation>
<category>A</category>
<remarks>zolt zolt</remarks>
</LP>
<LP>
<abbreviation>altalt</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>betbet</abbreviation>
<category>C</category>
<remarks />
</LP>
</LINKED_PROPERTIES>
<description>Area under administration</description>
<code>1</code>
<relation_type />
</FEATURE>
<FEATURE>
<abbreviation>BUOY</abbreviation>
<LINKED_PROPERTIES>
<LP>
<abbreviation>zltinf</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>altalt</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>betbet</abbreviation>
<category>C</category>
<remarks />
</LP>
<LP>
<abbreviation>infinf</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>altalt</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>betbet</abbreviation>
<category>C</category>
<remarks />
</LP>
</LINKED_PROPERTIES>
<description>Buoy on water</description>
<code>18</code>
<relation_type />
</FEATURE>
<FEATURE>
<abbreviation>DRONE</abbreviation>
<LINKED_PROPERTIES>
<LP>
<abbreviation>zltinf</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>beltam</abbreviation>
<category>A</category>
<remarks />
</LP>
<LP>
<abbreviation>betbet</abbreviation>
<category>C</category>
<remarks />
</LP>
</LINKED_PROPERTIES>
<description>Drones Inland</description>
<code>2</code>
<relation_type />
</FEATURE>
</FEATURES>本文档包含两个级别的功能,属于一个标准。所以很明显,我不能修改XML文档。上面的数据是匿名数据。
我的XQuery如下所示。
<FEATURES xmlns="http://www.example.com/" xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
{
let $doc := doc("features.xml")
for $feature in $doc/FEATURES/FEATURE
order by $feature/abbreviation ascending
return
<FEATURE>
{
for $ft in $feature
return if ($ft/child::element/name() eq
"LINKED_PROPERTIES")then
<LINKED_PROPERTIES_XXXXXXXX>{
for $lp in $ft/LINKED_PROPERTIES/LP
order
by $lp/abbreviation ascending
return $lp
}
</LINKED_PROPERTIES_XXXXXXXX>
else $ft/*}
</FEATURE>
}
</FEATURES>我无法在子级获得适当的相等条件,因此无法在子级获得所需的排序。
出于几个原因,我不想回到DOM/XSL。请协助。
发布于 2017-01-23 16:50:04
内部for需要迭代特性的子项,否则将只有一个项要迭代。此外,在树的更下面,必须删除LINKED_PROPERTIES,因为$ft已经在该节点上,并且必须按原样返回其他子节点。最后,这类测试不太正确(必须是child::element())。
<FEATURES xmlns="http://www.example.com/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">{
let $doc := doc("features.xml")
for $feature in $doc/FEATURES/FEATURE
order by $feature/abbreviation ascending
return <FEATURE>{
for $ft-child in $feature/child::element()
return if ($ft-child/name() eq "LINKED_PROPERTIES")
then <LINKED_PROPERTIES_XXXXXXXX>{
for $lp in $ft-child/LP
order by $lp/abbreviation ascending
return $lp
}</LINKED_PROPERTIES_XXXXXXXX>
else $ft-child
}</FEATURE>
}</FEATURES>以上查询在oXygen中成功。
发布于 2017-01-23 16:48:44
如果你想测试FEATURE是否有chid元素LINKED_PROPERTIES:
return if ($ft/LINKED_PROPERTIES) then
<LINKED_PROPERTIES_XXXXXXXX>
{
.....
}
</LINKED_PROPERTIES_XXXXXXXX>您的XQuery尝试同样没有多大意义,它永远不会计算为true,因为它查找名为element ($ft/child::element)的子元素,并进一步测试元素名称是否等于"LINKED_PROPERTIES" (/name() eq "LINKED_PROPERTIES")。
https://stackoverflow.com/questions/41800672
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