从Android上的URL简单解析JSON并在listview中显示

Question

我正在尝试解析从我的Android应用程序中的URL获取的JSON结果...

我在互联网上尝试了一些例子,但无法让它发挥作用.JSON数据如下所示:

[
    {
        "city_id": "1",
        "city_name": "Noida"
    },
    {
        "city_id": "2",
        "city_name": "Delhi"
    },
    {
        "city_id": "3",
        "city_name": "Gaziyabad"
    },
    {
        "city_id": "4",
        "city_name": "Gurgaon"
    },
    {
        "city_id": "5",
        "city_name": "Gr. Noida"
    }
]

获取URL并解析JSON数据的最简单方法是在列表视图中显示它

Answer 1

您可以使用AsyncTask,您必须自定义以满足您的需求,但类似于以下内容

异步任务有三种主要方法:

1. onPreExecute() - 最常用于设置和启动进度对话框

2. doInBackground() - 建立连接并从服务器接收响应(不要尝试将响应值分配给GUI元素,这是一个常见错误,无法在后台线程中完成).

3. onPostExecute() - 这里我们不在后台线程中,因此我们可以使用响应数据进行用户界面操作,或者只是将响应分配给特定的变量类型.

首先,我们将启动该类,初始化a String以将结果保存在方法之外但在类中,然后运行onPreExecute()方法设置一个简单的进度对话框.

class MyAsyncTask extends AsyncTask<String, String, Void> {

    private ProgressDialog progressDialog = new ProgressDialog(MainActivity.this);
    InputStream inputStream = null;
    String result = ""; 

    protected void onPreExecute() {
        progressDialog.setMessage("Downloading your data...");
        progressDialog.show();
        progressDialog.setOnCancelListener(new OnCancelListener() {
            public void onCancel(DialogInterface arg0) {
                MyAsyncTask.this.cancel(true);
            }
        });
    }

然后我们需要建立连接以及我们如何处理响应:

    @Override
    protected Void doInBackground(String... params) {

        String url_select = "http://yoururlhere.com";

        ArrayList<NameValuePair> param = new ArrayList<NameValuePair>();

        try {
            // Set up HTTP post

            // HttpClient is more then less deprecated. Need to change to URLConnection
            HttpClient httpClient = new DefaultHttpClient();

            HttpPost httpPost = new HttpPost(url_select);
            httpPost.setEntity(new UrlEncodedFormEntity(param));
            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();

            // Read content & Log
            inputStream = httpEntity.getContent();
        } catch (UnsupportedEncodingException e1) {
            Log.e("UnsupportedEncodingException", e1.toString());
            e1.printStackTrace();
        } catch (ClientProtocolException e2) {
            Log.e("ClientProtocolException", e2.toString());
            e2.printStackTrace();
        } catch (IllegalStateException e3) {
            Log.e("IllegalStateException", e3.toString());
            e3.printStackTrace();
        } catch (IOException e4) {
            Log.e("IOException", e4.toString());
            e4.printStackTrace();
        }
        // Convert response to string using String Builder
        try {
            BufferedReader bReader = new BufferedReader(new InputStreamReader(inputStream, "utf-8"), 8);
            StringBuilder sBuilder = new StringBuilder();

            String line = null;
            while ((line = bReader.readLine()) != null) {
                sBuilder.append(line + "\n");
            }

            inputStream.close();
            result = sBuilder.toString();

        } catch (Exception e) {
            Log.e("StringBuilding & BufferedReader", "Error converting result " + e.toString());
        }
    } // protected Void doInBackground(String... params)

最后,这里我们将解析返回,在这个例子中它是一个JSON数组然后关闭对话框:

    protected void onPostExecute(Void v) {
        //parse JSON data
        try {
            JSONArray jArray = new JSONArray(result);    
            for(i=0; i < jArray.length(); i++) {

                JSONObject jObject = jArray.getJSONObject(i);

                String name = jObject.getString("name");
                String tab1_text = jObject.getString("tab1_text");
                int active = jObject.getInt("active");

            } // End Loop
            this.progressDialog.dismiss();
        } catch (JSONException e) {
            Log.e("JSONException", "Error: " + e.toString());
        } // catch (JSONException e)
    } // protected void onPostExecute(Void v)
} //class MyAsyncTask extends AsyncTask<String, String, Void>

Answer 2

我建议使用这JSONParser门课.它非常易于使用.

public class JSONParser {

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

// constructor
public JSONParser() {

}

// function get json from url
// by making HTTP POST or GET method
public JSONObject makeHttpRequest(String url, String method,
        List<NameValuePair> params) throws IOException {

    // Making HTTP request
    try {

        // check for request method
        if(method == "POST"){
            // request method is POST
            // defaultHttpClient
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);
            httpPost.setEntity(new UrlEncodedFormEntity(params));

            HttpResponse httpResponse = httpClient.execute(httpPost);

            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();

        }else if(method == "GET"){
            // request method is GET
            DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
            String paramString = URLEncodedUtils.format(params, "utf-8");
            url += "?" + paramString;
            HttpGet httpGet = new HttpGet(url);

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


    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (Exception ex) {
        Log.d("Networking", ex.getLocalizedMessage());
        throw new IOException("Error connecting");
    }

    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;

}

然后在您的应用程序中,创建此类的实例.如果需要,您可能希望传递构造函数"GET"或"POST".

public JSONParser jsonParser = new JSONParser();

try {

    // Building Parameters ( you can pass as many parameters as you want)
    List<NameValuePair> params = new ArrayList<NameValuePair>();

    params.add(new BasicNameValuePair("name", name));
    params.add(new BasicNameValuePair("age", 25));

    // Getting JSON Object
    JSONObject json = jsonParser.makeHttpRequest(YOUR_URL, "POST", params);
} catch (JSONException e) {
    e.printStackTrace();
}

Answer 3

JSONObject(html).getString("name");

如何获取htmlString: 使用android发出HTTP请求