我有一张像这样的地图:
{
facility-1={
facility-kind1={param1=XPath-1, param2=XPath-2},
facility-kind2={param1=XPath-1, param2=XPath-2},
facility-kind3={param1=XPath-1, param2=XPath-2}
},
facility-2={
facility-kind1={param1=XPath-1, param2=XPath-2},
facility-kind2={param1=XPath-1, param2=XPath-2},
facility-kind3={param1=XPath-1, param2=XPath-2}
}
}我想把它转换成JSON格式,如下所示
[
{"title": "Item 1"},
{"title": "Folder 2",
"children": [
{"title": "Sub-item 2.1"},
{"title": "Sub-item 2.2"}
]
},
{"title": "Folder 3",
"children": [
{"title": "Sub-item 3.1"},
{"title": "Sub-item 3.2"}
]
},
{"title": "Item 5"}
]我尝试使用GSON,但得到的输出并不是我想要的:
{
"facility-1": {
"facility-kind1":
{"param1":"XPath-1","param2":"XPath-2"},
"facility-kind2":
{"param1":"XPath-1","param2":"XPath-2"},
"facility-kind3":
{"param1":"XPath-1","param2":"XPath-2"}
},
"facility-2": {
"facility-kind1":
{"param1":"XPath-1","param2":"XPath-2"},
"facility-kind2":
{"param1":"XPath-1","param2":"XPath-2","param3":"XPath-3"},
"facility-kind3":
{"param1":"XPath-1","param2":"XPath-2"}
}
}怎样才能得到我想要的json格式??
发布于 2012-06-07 14:55:40
您需要将JSON转换为您提供的新格式。
要转换的数据:
static String json =
"{\n" +
" facility-1={\n" +
" facility-kind1={param1=XPath-1, param2=XPath-2},\n" +
" facility-kind2={param1=XPath-1, param2=XPath-2},\n" +
" facility-kind3={param1=XPath-1, param2=XPath-2}\n" +
" },\n" +
" facility-2={\n" +
" facility-kind1={param1=XPath-1, param2=XPath-2},\n" +
" facility-kind2={param1=XPath-1, param2=XPath-2},\n" +
" facility-kind3={param1=XPath-1, param2=XPath-2}\n" +
" }\n" +
"}\n";使用GSON
创建一些您想要处理数据的类,它们可能如下所示:
static class Facility {
List<Kind> children = new LinkedList<Kind>();
}
static class Kind {
String title;
Map<String, String> params;
public Kind(String title, Map<String, String> params) {
this.title = title;
this.params = params;
}
}下一步也是看看源代码,并创建它的代表。我将使用:
Map<String, Map<String, Map<String, String>>>因为输入数据的布局与之类似。现在使用Gson转换它非常简单:
public static void main(String... args) throws Exception {
Gson gson = new GsonBuilder().setPrettyPrinting().create();
Type type = new TypeToken<
Map<String, Map<String, Map<String, String>>>>() {}.getType();
Map<String, Map<String, Map<String, String>>> source =
gson.fromJson(json, type);
Map<String, Facility> dest = new HashMap<String, Facility>();
for (String facilityName : source.keySet()) {
Map<String, Map<String, String>> facility = source.get(facilityName);
Facility f = new Facility();
for (String kindName : facility.keySet())
f.children.add(new Kind(kindName, facility.get(kindName)));
dest.put(facilityName, f);
}
System.out.println(gson.toJson(dest));
}使用JSONObject/JSONArray
public static void main(String... args) throws Exception {
JSONObject source = new JSONObject(json);
JSONArray destination = new JSONArray();
for (Iterator<?> keys = source.keys(); keys.hasNext(); ) {
String facilityName = (String) keys.next();
JSONObject kinds = source.getJSONObject(facilityName);
JSONArray children = new JSONArray();
for (Iterator<?> kit = kinds.keys(); kit.hasNext(); ) {
String kind = (String) kit.next();
JSONObject params = kinds.getJSONObject(kind);
JSONObject kindObject = new JSONObject();
kindObject.put("title", kind);
for (Iterator<?> pit = params.keys(); pit.hasNext(); ) {
String param = (String) pit.next();
kindObject.put(param, params.get(param));
}
children.put(kindObject);
}
JSONObject facility = new JSONObject();
facility.put("title", facilityName);
facility.put("children", children);
destination.put(facility);
}
System.out.println(destination.toString(2));
}发布于 2012-06-07 12:58:25
你想要的是漂亮的打印效果。
Gson gson = new GsonBuilder().setPrettyPrinting().create();
String json = gson.toJson(yourMap);https://stackoverflow.com/questions/10925874
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