JSON 解析为 Java - Android 应用程序

我需要在 Java Android 应用程序中解析 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，迈克尔。

使用手动解析，您可以像这样实现它：

            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");
}

在您的情况下，请注意您的外部元素数据是 JSONObject 类型，然后您有一个 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，但你的从 { 开始，然后里面有 [ 数组。如果您使用调试器运行它，您可以准确地看到 json 对象内部的内容。

还有更好的方法，例如：

1. Jackson http://wiki.fasterxml.com/JacksonInFiveMinutes
2. 杰克逊-JR https://github.com/FasterXML/jackson-jr（轻量级杰克逊）
3. GSON https://sites.google.com/site/gson/gson-user-guide

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