JSON解析为Java-Android应用程序

一尘不染

JSON解析为Java-Android应用程序

json

我需要在Java Android Appl中解析json字符串的帮助。

JSON文件的文本：

{"data":{"columns":["location_id","name","description","latitude","longitude","error","type","type_id","icon_media_id","item_qty","hidden","force_view"],"rows":[[2,"Editor","",43.076014654537,-89.399642451567,25,"Npc",1,0,1,"0","0"],[3,"Dow Recruiter","",43.07550842555,-89.399381822662,25,"Npc",2,0,1,"0","0"] [4,"Protestor","",43.074933,-89.400438,25,"Npc",3,0,1,"0","0"],[5,"State Legislator","",43.074868061524,-89.402136196317,25,"Npc",4,0,1,"0","0"],[6,"Marchers Bascom","",43.075296413877,-89.403374183615,25,"Node",22,0,1,"0","0"] [7,"Mary","",43.074997865584,-89.404967573966,25,"Npc",7,0,1,"0","0"]]},"returnCode":0,"returnCodeDescription":null}

如何获取值：location_id，名称，纬度，经度。谢谢，米哈尔。

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2020-07-27

共1个答案

一尘不染

使用手动解析，您可以这样实现：

            JSONArray  pages     =  new JSONArray(jsonString);
            for (int i = 0; i < pages.length(); ++i) {
                JSONObject rec = pages.getJSONObject(i);
                JSONObject jsonPage =rec.getJSONObject("page");
                String address = jsonPage.getString("url");
                String name = jsonPage.getString("name");
                String status =  jsonPage.getString("status");
}

在您的案例中，请注意您的外部元素数据是JSONObject类型，然后您有了一个JSONArray

我的json文件：

[{"page":{"created_at":"2011-07-04T12:01:00Z","id":1,"name":"Unknown Page","ping_at":"2011-07-04T12:06:00Z","status":"up","updated_at":"2011-07-04T12:01:00Z","url":"http://www.iana.org/domains/example/","user_id":2}},{"page":{"created_at":"2011-07-04T12:01:03Z","id":3,"name":"Down Page","ping_at":"2011-07-04T12:06:03Z","status":"up","updated_at":"2011-07-04T12:01:03Z","url":"http://www.iana.org/domains/example/","user_id":2}}]

请注意，我的从[开始，这意味着一个数组，但您的是从{开始，然后您在内部有[数组。如果使用调试器运行它，则可以准确看到json对象中的内容。

还有更好的方法，例如：

Jackson
Jackson-JR（轻量级Jackson）
GSON

所有这些都可以用于将Java对象转换为JSON表示形式。它还可以用于将JSON字符串转换为等效的Java对象。

2020-07-27