1

我正在连接一个网站及其 api 来检索数据。我下面的代码会执行此操作并获取响应正文，但是如何解析该响应正文？我是否必须创建自己的函数来搜索我想要的术语，然后获取每个术语的子内容？还是已经有一个我可以使用的库可以为我做到这一点？

private class GetResultTask extends AsyncTask<String, Void, String> {
    @Override
    protected String doInBackground(String... urls) {
      String response = "";

        DefaultHttpClient client = new DefaultHttpClient();
        HttpGet httpGet = new HttpGet("https://api.bob.com/2.0/shelves/45?client_id=no&whitespace=1");
        try {
          HttpResponse execute = client.execute(httpGet);
          InputStream content = execute.getEntity().getContent();

          BufferedReader buffer = new BufferedReader(new InputStreamReader(content));
          String s = "";
          while ((s = buffer.readLine()) != null) {
            response += s;
          }

        } catch (Exception e) {
          e.printStackTrace();
        }

      return response;
    }

    @Override
    protected void onPostExecute(String result) {
      apiStatus.setText(result); //setting the result in an EditText
    }
  }

响应体

{
"id": 415,
"url": "http://bob.com/45/us-state-capitals-flash-cards/",
"title": "U.S. State Capitals",
"created_by": "bub12",
"term_count": 50,
"created_date": 1144296408,
"modified_date": 1363506943,
"has_images": false,
"subjects": [
    "unitedstates",
    "states",
    "geography",
    "capitals"
],
"visibility": "public",
"has_access": true,
"description": "List of the U.S. states and their capitals",
"has_discussion": true,
"lang_terms": "en",
"lang_definitions": "en",
"creator": {
    "username": "bub12",
    "account_type": "plus",
    "profile_image": "https://jdfkls.dfsldj.jpg"
},
"terms": [
    {
        "id": 15407,
        "term": "Montgomery",
        "definition": "Alabama",
        "image": null
    },
    {
        "id": 15455,
        "term": "Juneau",
        "definition": "Alaska",
        "image": null
    },

    {
        "id": 413281851,
        "term": "Tallahassee",
        "definition": "Florida",
        "image": null
    },
    {
        "id": 413281852,
        "term": "Atlanta",
        "definition": "Georgia",
        "image": null
    }
  ]
}
4

4 回答 4

3

/ 那是 JSON，你可以使用 Jackson r Gson 之类的库来反序列化它。

http://jackson.codehaus.org/ http://code.google.com/p/google-gson/

您可以将 Java 对象映射到 Json 或像通用对象一样反序列化它。

于 2013-05-25T00:40:38.137 回答
3

该数据格式是 JSON（JavaScript Object Notation）。因此，您只需要一个与 android 兼容的 JSON 解析器，例如GSON，就可以开始了。

于 2013-05-25T00:39:00.820 回答
2

SpringRestTemplate非常简单，它会自动将响应主体直接解组（即解析）为与响应的 JSON 格式匹配的 Java 对象：

首先，定义 Java 类以匹配数据格式，必要时使用 JAXB 注释。这是一个基于您的响应主体的稍微简化的模型：

@XmlRootElement
class MyResponseData {
    long id;
    String url;
    String title;
    String created_by;
    int term_count;
    int created_date;
    int modified_date;
    boolean has_images;
    List<String> subjects;
    Creator creator;
    List<Term> terms;
}

class Creator {
    String username;
    String account_type;
    String profile_image;
}

class Term {
    long id;
    String term;
    String definition;
    String image;
}

然后你只需用 Spring 的RestTemplate

String url = "https://api.bob.com/2.0/shelves/45?client_id=no&whitespace=1";
RestTemplate template = new RestTemplate();
MyResponseData body = template.getForObject(url, MyResponseData.class);

3 行代码发出请求并将响应主体作为 Java 对象获取。它并没有变得更简单。

http://static.springsource.org/spring/docs/current/javadoc-api/org/springframework/web/client/RestTemplate.html

于 2013-05-25T02:21:46.213 回答
0

在您的包中添加以下 JSONParser.java 类并使用，

你的类.java

JSONObject json = jParser.getJSONFromUrl(YOUR_URL_TO_JSON);

try {
        // Getting Array of terms
    JSONArray terms = json.getJSONArray("terms");

    // looping through All Contacts
    for(int i = 0; i < terms.length; i++){

         JSONObject termsJsonObject= terms.getJSONObject(i);

             String id = termsJsonObject.getJSONObject("id").toString();
             String term = termsJsonObject.getJSONObject("term").toString();
             String definition = termsJsonObject.getJSONObject("definition").toString();
             String image = termsJsonObject.getJSONObject("image").toString();

             // do  your operations using these values
         }
  }
  catch (JSONException e) {
        e.printStackTrace();
    return "";
  }

JSONParser.java

public class JSONParser {

    static InputStream is = null;
    static JSONObject jObj = null;
    static String json = "";

    // constructor
    public JSONParser() {

    }

    public JSONObject getJSONFromUrl(String url) {

        // Making HTTP request
        try {
            // defaultHttpClient
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);

            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();          

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            json = sb.toString();
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }

        // try parse the string to a JSON object
        try {
            jObj = new JSONObject(json);
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }

        // return JSON String
        return jObj;

    }
}
于 2013-05-25T02:00:10.900 回答