构造JSON(Constructing JSON)

构造JSON(Constructing JSON)

我的文档是XML格式，我需要根据这个文档构建一个JSON格式。 首先，我创建了一个返回正确XML结构的查询，但是当我将此查询转换为返回JSON时，我遇到了For Return问题。

Error: [1.0-ml] XDMP-UNDVAR: (err:XPST0008) Undefined variable $cat

---

原始查询：

let $tree := <tree>{for $cat in distinct-values($doc//ns:category/@name) return <cat name="{$cat}"> { for $var in $doc//ns:category[@name = $cat]//ns:variable/@name return <var name="{$var}"> } </var> } </cat> } </tree> return $tree

---

数据：

<category name="Catname"> <variable name="Varname"> <segment name="Seg1">9</segment> <segment name="Seg2">33</segment> <segment name="Seg3">32</segment> <segment name="Seg4">22</segment> </variable> <variable name="Vartwo"> <segment name="Seg2one">1</segment> <segment name="Seg2two">2</segment> </variable> </category> <category> ....

---

新查询：

let $tree := json:array( <json:array xmlns:json="http://marklogic.com/xdmp/json" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <json:value xsi:type="xs:string"> {for $cat in distinct-values($doc//ns:category/@name) return $cat} </json:value> <json:array> {for $var in $doc//ns:category[@name = $cat]//ns:variable/@name return $var} </json:array> </json:array> ) return $tree

---

预期产出

{"Catname": ["Varname": ["Seg1", "Seg2", "Seg3"], "Vartwo": ["Seg2one", "Seg2two"]]}, {"Cat2" ... }

My document is in XML and I need to construct a JSON format based on this document. First I created a query returning the right XML structure but when I convert this query to return JSON, I have a For Return problem.

Error: [1.0-ml] XDMP-UNDVAR: (err:XPST0008) Undefined variable $cat

---

Original query:

let $tree := <tree>{for $cat in distinct-values($doc//ns:category/@name) return <cat name="{$cat}"> { for $var in $doc//ns:category[@name = $cat]//ns:variable/@name return <var name="{$var}"> } </var> } </cat> } </tree> return $tree

---

Data:

<category name="Catname"> <variable name="Varname"> <segment name="Seg1">9</segment> <segment name="Seg2">33</segment> <segment name="Seg3">32</segment> <segment name="Seg4">22</segment> </variable> <variable name="Vartwo"> <segment name="Seg2one">1</segment> <segment name="Seg2two">2</segment> </variable> </category> <category> ....

---

New Query:

let $tree := json:array( <json:array xmlns:json="http://marklogic.com/xdmp/json" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <json:value xsi:type="xs:string"> {for $cat in distinct-values($doc//ns:category/@name) return $cat} </json:value> <json:array> {for $var in $doc//ns:category[@name = $cat]//ns:variable/@name return $var} </json:array> </json:array> ) return $tree

---

Expected Output

{"Catname": ["Varname": ["Seg1", "Seg2", "Seg3"], "Vartwo": ["Seg2one", "Seg2two"]]}, {"Cat2" ... }

最满意答案

您收到该错误是因为变量$cat不在嵌套<json:array/>元素的范围内。 它在前一个范围中定义，因此不可用。

我不确定你想要创建什么JSON结构，但是 使用XML序列化来创建JSON肯定很尴尬。 我更喜欢这样的方法：

更新：提供的JSON结构无效; 我冒昧地将类别属性转换为对象。 这些XQuery示例将输出JSON，如下所示：

{"Catname":{"Vartwo":["Seg2one","Seg2two"], "Varname":["Seg1","Seg2","Seg3","Seg4"]}}

此方法使用功能映射API：

let $map := map:new( for $cat in distinct-values($doc//ns:category/@name) return map:entry($cat, for $var in $doc//ns:category[@name = $cat]//ns:variable let $name := $var/@name/fn:string() return map:entry($name, json:to-array($var/ns:segment/@name/fn:string())) )) return xdmp:to-json($map)

以下是使用过程映射API的相同代码：

let $map := map:map() let $_ := for $cat in distinct-values($doc//ns:category/@name) let $cat-map := map:map() let $_ := for $var in $doc//ns:category[@name = $cat]//ns:variable let $name := $var/@name/fn:string() return map:put($cat-map, $name, json:to-array($var/ns:segment/@name/fn:string())) return map:put($map, $cat, $cat-map) return xdmp:to-json($map)

这种方法的关键点是

使用json：to-array（）从XQuery序列构造数组 使用xdmp：to-json（）从地图构造对象

此代码从您的示例XML返回以下JSON： {"Catname":["Varname"]} ，这似乎不太有用。 一旦添加了想要查看的输出，我就会更新。

You're getting that error because the variable $cat is not in scope inside of the nested <json:array/> element. It's defined in the previous scope, and, therefore, not available.

I'm not sure what JSON structure you want to create, but it's definitely awkward to use that XML serialization to create JSON. I prefer an approach like this:

Update: The provided JSON structure isn't valid; I've taken the liberty of converting the category property to an object. These XQuery samples will output JSON as follows:

{"Catname":{"Vartwo":["Seg2one","Seg2two"], "Varname":["Seg1","Seg2","Seg3","Seg4"]}}

This approach uses the functional maps API:

let $map := map:new( for $cat in distinct-values($doc//ns:category/@name) return map:entry($cat, for $var in $doc//ns:category[@name = $cat]//ns:variable let $name := $var/@name/fn:string() return map:entry($name, json:to-array($var/ns:segment/@name/fn:string())) )) return xdmp:to-json($map)

And here's the same code, using the procedural maps API:

let $map := map:map() let $_ := for $cat in distinct-values($doc//ns:category/@name) let $cat-map := map:map() let $_ := for $var in $doc//ns:category[@name = $cat]//ns:variable let $name := $var/@name/fn:string() return map:put($cat-map, $name, json:to-array($var/ns:segment/@name/fn:string())) return map:put($map, $cat, $cat-map) return xdmp:to-json($map)

The key takeaways from this approach are

construct arrays from XQuery sequences using json:to-array() construct objects from maps using xdmp:to-json()

This code returns the following JSON from your sample XML: {"Catname":["Varname"]}, which doesn't seem very useful. I'll update once you've added the output you want to see.