(BaseX) 错误

我在 BaseX 中对一个大型数据集运行查询，但一个 XQuery 使我的程序崩溃，错误 [XPTY0004] Item expected, sequence found: (attribute begin {"6"}, ...). .

在我的查询中，我试图通过比较 begin 来确保一个元素排在另一个元素之前。 - 存在于 XML 中的属性 - 带有 number() .但是每当我在我的数据集上尝试最基本的 XQueries(返回匹配节点)(例如使用 this online tool )时，我都会收到一个类似于我之前遇到的错误:

[Error] SaxonCE.XSLT20Processor 14:08:39.692 SEVERE: XPathException in invokeTransform: A sequence of more than one item is not allowed as the first argument of number() ("6", "10")

node[@cat="np" and node[@pt="art"] and node[@pt="adj"] and node[@pt="n"]]

# Possible representation of *the time available*
<node id="0" begin="1" cat="np">
    <node id="1" begin="1" pt="art" text="the" />        
    <node id="2" begin="2" pt="n" text="time" />
    <node id="3" begin="3" pt="adj" text="available" />
</node>

# Possible representation of *the concerned man*
<node id="0" begin="1" cat="np">
    <node id="1" begin="1" pt="art" text="the" />        
    <node id="2" begin="2" pt="adj" text="concerned" />
    <node id="3" begin="3" pt="n" text="man" />
</node>

<node id="0" begin="2">
    <node id="1" begin="2">
        <node id="2" begin="2"/>
        <node id="3" begin="3"/>
    </node>
    <node id="4" begin="5">
        <node id="5" begin="5"/>
    </node>
    <node id="6" begin="6"/>
</node>

/node/node[number(@begin) < number(../node/@begin)]

for $node in node[every $n in node[position() lt last()] satisfies (number($n/@begin) lt number($n/following-sibling::node/@begin))]
return $node

<node id="0" begin="2">
    <node id="1" begin="2">
        <node id="2" begin="2"/>
        <node id="3" begin="3"/>
    </node>
    <node id="4" begin="5">
        <node id="5" begin="5"/>
    </node>
    <node id="6" begin="6"/>
</node>

SaxonCE.XSLT20Processor 14:48:49.809 SEVERE: XPathException in invokeTransform: A sequence of more than one item is not allowed as the first argument of number() ("5", "6")

//node[@cat="np" and node[@rel="det" and @pt="vnw" and @lemma="die"] and count(node[@rel="mod" and @pt="adj"]) > 1]

//node[
    @cat="np" and 
    not(node[
        position() < last()
    ][number(@begin) > following-sibling::node/number(@begin)]) and 
    node[
        @rel="det" and 
        @pt="vnw" and 
        @lemma="die"
    ] and 
    count(node[
        @rel="mod" and 
        @pt="adj"
    ]) > 1
]

<node begin="4" cat="np" id="8" rel="obj1">
    <node begin="4" id="9" pos="det" pt="vnw" rel="det" word="die" lemma="die" />
    <node begin="5" id="10" pos="adj" pt="adj" rel="mod" word="veelzijdige" />
    <node begin="6" id="11" pos="adj" pt="adj" rel="mod" word="getalenteerde" />
    <node begin="7" id="12" pos="noun" pt="n" rel="hd" word="figuren" />
    <node begin="8" id="31" index="1" rel="obj1" />
    <node begin="2" id="32" index="2" rel="obj2" />
</node>

最佳答案

我认为我理解的问题部分是这样的:

假设我想匹配 XML，其中根的每个直接子级的属性 begin 都小于下一个兄弟级。

<node id="0" begin="2">
    <node id="1" begin="2">
        <node id="2" begin="2"/>
        <node id="3" begin="3"/>
    </node>
    <node id="4" begin="5">
        <node id="5" begin="5"/>
    </node>
    <node id="6" begin="6"/>
</node>

/node/node[number(@begin) < number(../node/@begin)]

node[every $n in node[position() lt last()] satisfies (number($n/@begin) lt number($n/following-sibling::node/@begin)]

关于xml - XPath 拉出不止一场比赛，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42673237/