我发现以下代码适用于我的一小部分数据，但我没有意识到我没有采用任何带有多个注释的样本。当我尝试将代码应用于实际数据库时，每个条目有多个注释，我收到了上面提到的错误。
当前代码：
for $doc in doc('test')
let $results :=
(
let $pKeywords := ('best clients', 'Very', '20')
return
for $kw in $pKeywords
return
(
$doc/set/entry[contains(comment, concat('!', $kw))],
$doc/set/entry[contains(comment, $kw)]
)
[not(position() gt 2)]
)
for $i in (1 to count($results))
return
(
subsequence($results/comment, $i, 1),
subsequence($results/buyer, $i, 1)
)
文件：
<set>
<entry>
<comment>The client is only 20 years old. Do not be surprised by his youth.</comment>
<buyer></buyer>
<id>1282</id>
<industry>International Trade; Fish and Game</industry>
</entry>
<entry>
<comment>!On leave in October.</comment>
<comment>!Planning to make a large purchase before Christmas.</comment>
<buyer></buyer>
<id>709</id>
<industry>Real Estate</industry>
</entry>
<entry>
<comment>Is often !out between 1 and 3 p.m.</comment>
<buyer></buyer>
<id>127</id>
<industry>Virus Software Marketting</industry>
</entry>
<entry>
<comment>Very personable. One of our best clients.</comment>
<buyer></buyer>
<id>14851</id>
<industry>Administrative support.</industry>
</entry>
<entry>
<comment>!Very difficult to reach, but one of our top buyers.</comment>
<comment>His wife often answers the phone. That means he is out of the office.</comment>
<buyer></buyer>
<id>1458</id>
<industry>Construction</industry>
</entry>
<entry>
<comment></comment>
<buyer></buyer>
<id>276470</id>
<industry>Bulk Furniture Sales</industry>
</entry>
<entry>
<comment>A bit of an eccentric. One of our best clients.</comment>
<buyer></buyer>
<id>1506</id>
<industry>Sports Analysis</industry>
</entry>
<entry>
<comment>Very gullible, so please !be sure she needs what you sell her. She's one of our best clients.</comment>
<buyer></buyer>
<id>1523</id>
<industry>International Trade</industry>
</entry>
<entry>
<comment>He wants to buy everything, but !he has a tight budget.</comment>
<comment>!His company may be closing soon.</comment>
<buyer></buyer>
<id>1524</id>
<industry>Public Relations</industry>
</entry>
</set>
结果：
Stopped at line 9, column 22: [XPTY0004] document-node()(...): function(item()*) as item()* expected, document-node() found.
我遇到了类似的错误并且能够修复它，但是当我尝试应用修复时，这不起作用。例如：
$doc('test')/set/entry[contains(., concat('!', $kw))],
$doc('test')/set/entry[contains(., $kw)]
返回相同的结果。
完成所需的结果：
如果条目return子项包含entry中的三个关键字中的任何一个，则第一个comment应返回每个$pKeywords及其子级。
concat('!', $kw)应该使！-containing comments成为优先事项。
第二个return对第一个comment的结果中的buyer和return个节点进行切片。
只要每个条目中只有1个comment - 命名节点，代码就可以正常运行。当有2个或更多comment个命名节点时，代码失败，编译器返回上述错误：
Stopped at line 9, column 22: [XPTY0004] document-node()(...): function(item()*) as item()* expected, document-node() found.
- 编辑 -
期望的结果：
<comment>The client is only 20 years old. Do not be surprised by his youth.</comment>
<buyer/>
<comment>Very personable. One of our best clients.</comment>
<buyer/>
<comment>!Very difficult to reach, but one of our top buyers.</comment>
<buyer/>
<comment>A bit of an eccentric. One of our best clients.</comment>
<buyer/>
澄清所需的结果：
//contains ! and the first keyword, "best clients"; so, the first result should come from this entry.
<entry>
<comment>Very gullible, so please !be sure she needs what you sell her. She's one of our best clients.</comment>
<buyer></buyer>
<id>1523</id>
<industry>International Trade</industry>
</entry>
//Only one entry contains ! and "best clients". So, the first result containing "best clients" contains nodes for the second result.
<entry>
<comment>Very personable. One of our best clients.</comment>
<buyer></buyer>
<id>14851</id>
<industry>Administrative support.</industry>
</entry>
//This contains ! and the second keyword, "Very", but it is a duplicate. So, ideally its children should not be returned.
<entry>
<comment>!Very difficult to reach, but one of our top buyers.</comment>
<comment>His wife often answers the phone. That means he is out of the office.</comment>
<buyer></buyer>
<id>1458</id>
<industry>Construction</industry>
</entry>
//This contains ! and a string, "very" (part of everything). Nodes from this entry should be returned as the third result.
<entry>
<comment>He wants to buy everything, but !he has a tight budget.</comment>
<comment>!His company may be closing soon.</comment>
<buyer></buyer>
<id>1524</id>
<industry>Public Relations</industry>
</entry>
//The only entry whose comment child contains the keyword '20'. There is no '!'-containing comment with 20, so this nodes is the top and only node whose children should be returned.
<entry>
<comment>The client is only 20 years old. Do not be surprised by his youth.</comment>
<buyer></buyer>
<id>1282</id>
<industry>International Trade; Fish and Game</industry>
</entry>
-Edit 2 -
下一步可以更好地了解我想要完成的任务，但是有一些明显的语法错误（例如，我仍然在发现如何使用数组，如第8行所示）。我将在解决语法错误时对此进行更新：
<set>
{
let $kw := ('best clients', 'Very', '20')
let $entry := doc('test')/set/entry
let $priority := '!'
for $i in (1, count($kw))
let $priority_result[$i] :=
(
for $entries in $entry
where $entry contains(., $priority) and where $entry contains $kw[$i]
return subsequence($priority_result[$i], 1, 2)
)
if $priority_result[$i] < 2
for $i in (1, count($kw))
let $secondary_result[$i] :=
(
for $entries in $entry
where $entry contains $kw[$i] and where $entry not($priority_result) and where $entry not($secondary_result[1..($i-1)])
return $secondary_result[$i]
)
else let $secondary_result[$i] := ''
for $i in (1, count($kw))
return
(
$primary_result[$i],
$secondary_result[$i]
)
}
</set>
建议的更改，返回null结果：
for $doc in doc('test')
let $results :=
(
let $pKeywords := ('best clients', 'Very', '20')
return
for $kw in $pKeywords
return
(
$doc/set/entry/comment[contains(., concat('!', $kw))],
$doc/set/entry/comment[contains(., $kw)]
)
[not(position() gt 2)]
)
for $i in (1 to count($results))
return
(
subsequence($results/comment, $i, 1),
subsequence($results/buyer, $i, 1)
)
答案 0 :(得分：1)
错误消息似乎在抱怨尝试将document-node()作为函数调用。
$doc('test') vs $doc
或者，comments(...)仅适用于单个节点，而不适用于节点集。
contains(comment, $kw) vs comment/contains(.,$kw)
或comment[contains(.,$kw)]
或comment[contains(text(),$kw)]
这对我有用：
<set>{
for $entry in doc('test')/set/entry
let $kw := (
for $prefix in ('!','')
for $kw in ('best clients', 'Very', '20')
where exists($entry/comment[contains(., concat($prefix,$kw))])
return concat($prefix,$kw)
)[1]
where exists($kw)
order by not(starts-with($kw,'!'))
return <entry keyword="{$kw}">{
( $entry/comment,
$entry/buyer )
}</entry>
}</set>
结果（每<entry>多个评论）：
<set>
<entry keyword="!Very">
<comment>!Very difficult to reach, but one of our top buyers.</comment>
<comment>His wife often answers the phone. That means he is out of the office.</comment>
<buyer/>
</entry>
<entry keyword="20">
<comment>The client is only 20 years old. Do not be surprised by his youth.</comment>
<buyer/>
</entry>
<entry keyword="best clients">
<comment>Very personable. One of our best clients.</comment>
<buyer/>
</entry>
<entry keyword="best clients">
<comment>A bit of an eccentric. One of our best clients.</comment>
<buyer/>
</entry>
<entry keyword="best clients">
<comment>Very gullible, so please !be sure she needs what you sell her. She's one of our best clients.</comment>
<buyer/>
</entry>
</set>
这将为每条评论提供单独的条目：
<set>{
for $entry in doc('test')/set/entry
for $comment in $entry/comment
let $kw := (
for $prefix in ('!','')
for $kw in ('best clients', 'Very', '20')
where exists($comment[contains(., concat($prefix,$kw))])
return concat($prefix,$kw)
)[1]
where exists($kw)
order by not(starts-with($kw,'!'))
return <entry keyword="{$kw}">{
( $comment,
$entry/buyer )
}</entry>
}</set>
输出：
<set>
<entry keyword="!Very">
<comment>!Very difficult to reach, but one of our top buyers.</comment>
<buyer/>
</entry>
<entry keyword="20">
<comment>The client is only 20 years old. Do not be surprised by his youth.</comment>
<buyer/>
</entry>
<entry keyword="best clients">
<comment>Very personable. One of our best clients.</comment>
<buyer/>
</entry>
<entry keyword="best clients">
<comment>A bit of an eccentric. One of our best clients.</comment>
<buyer/>
</entry>
<entry keyword="best clients">
<comment>Very gullible, so please !be sure she needs what you sell her. She's one of our best clients.</comment>
<buyer/>
</entry>
</set>
答案 1 :(得分：0)
为了参考起见，这是我们开始的代码（它有点令人生畏，我仍然不明白）：
for $doc in doc('test')
let $results :=
(
let $pKeywords := ('best clients', 'Very', '20')
return
for $kw in $pKeywords
return
(
$doc/set/entry[contains(comment, concat('!', $kw))], (: *1 :)
$doc/set/entry[contains(comment, $kw)] (: *1 :)
)
[not(position() gt 2)]
)
for $i in (1 to count($results))
return
(
subsequence($results/comment, $i, 1), (: *2 :)
subsequence($results/buyer, $i, 1) (: *2 :)
)
不会抛出错误的版本以典型方式解决。我花了一些时间来捕获第二个错误，标记为*2。基本上，因为我在搜索中更深入一级*1，所以我需要在结果上升一级，..：
for $doc in doc('test')
let $results :=
(
let $pKeywords := ('best clients', 'Very', '20')
return
for $kw in $pKeywords
return
(
$doc/set/entry/comment[contains(., concat('!', $kw))], (: *1, went deeper :)
$doc/set/entry/comment[contains(., $kw)] (: *1, went deeper :)
)
[not(position() gt 2)]
)
for $i in (1 to count($results))
return
(
subsequence($results/../comment, $i, 1), (: *2, added .. :)
subsequence($results/../buyer, $i, 1) (: *2, added .. :)
)
我还在努力：
1）使用concat()。我的理解是它将两个事物放在一起，$kw[1]的结果将等同于"!best clients"。但结果并没有显示出来。在结果中，感叹号并不总是直接位于优先级查询之前。
2）不返回重复结果。我希望每个条目都是独一无二的。我需要在某个地方添加一个例程，要么限制重复项进入我的结果集，要么消除[not(position() gt 2)]之前的重复项，其中结果数被修剪/切片。
感谢所有观众和努力工作！仍然期待更好的答案！