Question

所以我有一个JSONObject（或String ..），如下所示：

    {"locations":[{"GeocodeResponse":{"result":{"formatted_address":"Tchibanga (TCH), Gabon","address_component":[{"long_name":"Tchibanga","type":
["airport","establishment","transit_station"],"short_name":"Tchibanga"},{"long_name":"Mougoutsi","type":
["administrative_area_level_2","political"],"short_name":"Mougoutsi"},{"long_name":"Nyanga","type":
["administrative_area_level_1","political"],"short_name":"Nyanga"},{"long_name":"Gabon","type":
["country","political"],"short_name":"GA"}],"type":["airport","establishment","transit_station"],"geometry":
{"viewport":{"southwest":{"lng":"10.9968524","lat":"-2.8198146"},"northeast":{"lng":"11.0031476","lat":"-2.8135194"}},"location_type":"APPROXIMATE","location":{"lng":"11","lat":"-2.816667"}}},"status":"OK"}}]}

然而，信息太多了，我只想说

{"locations":[{"id":"Tchibanga(TCH)","parentId":"TCH","airport":"Tchibanga","category":"Airport","location":{"longitude":"11","latitude":"-2.816667"},"name":"Nyanga","country":"GA"}]}

我将如何正确地解析这个？

编辑：不，我不想让另一个图书馆解析它。

Answer 1

看来你想要的只是简单地将一块JSON转换成一个不同的块。也许这个问题可以帮到你：XSLT equivalent for JSON

Answer 2

检查杰克逊图书馆。 http://jackson.codehaus.org/

Answer 3

您需要从here获取JSON库（您必须编译源并确保类在您的类路径上）并创建JSONObject。

JSONObject只是一个包含更多地图，数组和对象的地图。解析很容易（但很麻烦）因为有很多嵌套。我们来看看如何解析第一个long_name的值。如果您查看JSON源字符串，您会看到long_name的位置位于locations/GeocodeResponse/result/address_component。所以你会做这样的事情：

//create a jsonObject
JSONObject jsonObject = new JSONObject("{ \"locations\" ...<snipped>... ] }");

//run some getters until you get to the address_component
JSONArray locations = (JSONArray)jsonObject.get("locations");
JSONObject location = (JSONObject)locations.get(0); // get the first location
JSONObject geoCodeResponse = (JSONObject)location.get("GeocodeResponse");
JSONObject result = (JSONObject)geoCodeResponse.get("result");
JSONArray addressArray = (JSONArray)result.get("address_component");

//print out the long_name from the address
JSONObject address = (JSONObject)addressArray.get(0);
String longName = (String)address.get("long_name");
System.out.println(longName); //prints Tchibanga

但是，我建议您使用JsonPath，以便让您的生活更轻松。

一旦你挑选出你需要的元素，你就可以构建所需的输出JSONObject。

Answer 4

Flexjson允许您在对象图中包含/排除特定对象。

用Java重新编写JSON对象？

4 个答案: