JSONPath在Java中的基本用法 [英] Basic use of JSONPath in Java

问题描述

我将JSON作为字符串，将JSONPath作为字符串。我想用JSON路径查询JSON，将生成的JSON作为字符串。

I have JSON as a string and a JSONPath as a string. I'd like to query the JSON with the JSON path, getting the resulting JSON as a string.

我收集 Jayway的json-path 是标准。 在线API但是，与你从Maven获得的实际图书馆。 GrepCode的版本大致匹配。

I gather that Jayway's json-path is the standard. The online API, however, doesn't have have much relation to the actual library you get from Maven. GrepCode's version roughly matches up though.

看起来我应该能够做到：

It seems like I ought to be able to do:

String originalJson; //these are initialized to actual data
String jsonPath;
String queriedJson = JsonPath.<String>read(originalJson, jsonPath);

问题是读取返回任何它根据JSONPath实际发现的内容感觉最合适（例如 List< Object> ， String ， double 等），因此我的代码会针对某些查询抛出异常。假设有一些方法可以查询JSON并获取JSON，这似乎是合理的。有什么建议吗？

The problem is that read returns whatever it feels most appropriate based on what the JSONPath actually finds (e.g. a List<Object>, String, double, etc.), thus my code throws an exception for certain queries. It seems pretty reasonable to assume that there'd be some way to query JSON and get JSON back; any suggestions?

推荐答案

肯定有一种方法可以查询Json并使用JsonPath获取Json。
参见下面的示例：

There definitely exists a way to query Json and get Json back using JsonPath. See example below:

 String jsonString = "{\"delivery_codes\": [{\"postal_code\": {\"district\": \"Ghaziabad\", \"pin\": 201001, \"pre_paid\": \"Y\", \"cash\": \"Y\", \"pickup\": \"Y\", \"repl\": \"N\", \"cod\": \"Y\", \"is_oda\": \"N\", \"sort_code\": \"GB\", \"state_code\": \"UP\"}}]}";
 String jsonExp = "$.delivery_codes";
 JsonNode pincodes = JsonPath.read(jsonExp, jsonString, JsonNode.class);
 System.out.println("pincodesJson : "+pincodes);

上面的输出将是内部Json。

The output of the above will be inner Json.

[{postal_code：{district：Ghaziabad，pin：201001，pre_paid：Y，cash：Y，pickup ： Y， REPL： N， 鳕鱼： Y， is_oda： N， sort_code： GB， STATE_CODE： UP}}]

[{"postal_code":{"district":"Ghaziabad","pin":201001,"pre_paid":"Y","cash":"Y","pickup":"Y","repl":"N","cod":"Y","is_oda":"N","sort_code":"GB","state_code":"UP"}}]

现在可以通过迭代我们上面得到的List（JsonNode）来解析每个单独的名称/值对。

Now each individual name/value pairs can be parsed by iterating the List (JsonNode) we got above.

for(int i = 0; i< pincodes.size();i++){
    JsonNode node = pincodes.get(i);
    String pin = JsonPath.read("$.postal_code.pin", node, String.class);
    String district = JsonPath.read("$.postal_code.district", node, String.class);
    System.out.println("pin :: " + pin + " district :: " + district );
}

输出将是：

pin :: 201001 district :: Ghaziabad

pin :: 201001 district :: Ghaziabad

取决于你想要的Json解析，您可以决定是获取List还是只获取一个String / Long值。

Depending upon the Json you are trying to parse, you can decide whether to fetch a List or just a single String/Long value.

希望它有助于解决您的问题。

Hope it helps in solving your problem.

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