json获取序言谓词 [英] json get prolog predicate

问题描述

我正在尝试在序言中创建此谓词:

I'm tryung to create this predicate in prolog:

谓词json_get/3可以定义为:json_get(JSON_obj, Fields, Result).，当Result可通过以下方式恢复时为true 从JSON_obj开始的Fields(列表)中的字段链.一个领域 用N表示(N主数o等于0)表示与 JSON数组的索引.

The predicate json_get/3 can be defined as: json_get(JSON_obj, Fields, Result). which is true when Result is recoverable by following the chain of fields in Fields (a list) starting from JSON_obj. A field represented by N (with N a major number o equal to 0) corresponds to an index of a JSON array.

请帮助我了解遵循字段链的方法.

Please help me to understand to follow the chain of fields.

谢谢

edit1:

当然，因此json对象看起来像这样的"{{name":"Aretha"，"surname":"Franklin"}". 如果我将此对象的序言调用json_parse谓词显示给我

Of course, so json object looks like this '{"name" : "Aretha", "surname" : "Franklin"}'. if i call json_parse predicate to this object prolog show me this

json_obj([(name, Aretha), (surname, Franklin)])，我们称为obj O.

json_obj([("name", "Aretha"), ("surname", "Franklin")]), we call this obj O.

使用json_get我需要以这种方式从O中提取名称json_get(O, ["name"], R)

with json_get i need to extract from O the name in this way, json_get(O, ["name"], R)

edit2:

在某人的帮助下，这是现在的谓词:

with someone's help this is the predicate now:

json_get(json_obj(JSON_obj), Field, Result) :-
    memberchk((Field,Result), JSON_obj).

json_get(JSON_obj, Fields, Result) :-
    maplist(json_get(JSON_obj), Fields, Result).

所以现在的问题是嵌套列表. 例如，使用此输入

so now the problem is nested list. For example with this input

json_parse('{"nome" : "Zaphod",
            "heads" : ["Head1", "Head2"]}', Z),
json_get(Z, ["heads", 1], R).

输出应为R = "Head2"，但谓词不会提取该字段并失败.

the output will should be R = "Head2" but the predicate doesn't extract the field and fail.

edit3:

这是json_parse的输出

this is the output of json_parse

json_obj([("nome", "Zaphod"),  ("heads", json_array(["Head1", "Head2"]))]).

推荐答案

如何解决

json_get(json_obj(Obj),[F|Fs],Res) :-
  member((F,R),Obj),
  json_get(R,Fs,Res).
json_get(json_array(Is),[N|Fs],Res) :-
  nth1(N,Is,R),
  json_get(R,Fs,Res).
json_get(Res,[],Res).

这将在您的第二个示例中生成Head1而不是Head2.如果您不只是打错字，请说明该怎么做. (如果它是从零开始的，则可以将nth1/3更改为nth0/3.)

This produces Head1 not Head2 in your 2nd example. Please explain how that is supposed to work, if you did not just make a typo. (If it is zero-based you can just change nth1/3 to nth0/3.)

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