如何在没有数组名称的情况下编写Android Json解析

Question

 package info.androidhive.jsonparsing;

import android.app.ProgressDialog;
import android.os.AsyncTask;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.util.Log;
import android.widget.ListAdapter;
import android.widget.ListView;
import android.widget.SimpleAdapter;
import android.widget.Toast;

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import java.util.ArrayList;
import java.util.HashMap;

public class MainActivity extends AppCompatActivity {

private String TAG = MainActivity.class.getSimpleName();

private ProgressDialog pDialog;
private ListView lv;

// URL to get contacts JSON
private static String url = "http://api.androidhive.info/contacts/";

ArrayList<HashMap<String, String>> contactList;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    contactList = new ArrayList<>();

    lv = (ListView) findViewById(R.id.list);

    new GetContacts().execute();
}

/**
 * Async task class to get json by making HTTP call
 */
private class GetContacts extends AsyncTask<Void, Void, Void> {

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        // Showing progress dialog
        pDialog = new ProgressDialog(MainActivity.this);
        pDialog.setMessage("Please wait...");
        pDialog.setCancelable(false);
        pDialog.show();

    }

    @Override
    protected Void doInBackground(Void... arg0) {
        HttpHandler sh = new HttpHandler();

        // Making a request to url and getting response
        String jsonStr = sh.makeServiceCall(url);

        Log.e(TAG, "Response from url: " + jsonStr);

        if (jsonStr != null) {
            try {
                JSONObject jsonObj = new JSONObject(jsonStr);

                // Getting JSON Array node
                JSONArray contacts = jsonObj.getJSONArray("contacts");

                // looping through All Contacts
                for (int i = 0; i < contacts.length(); i++) {
                    JSONObject c = contacts.getJSONObject(i);

                    String id = c.getString("id");
                    String name = c.getString("name");
                    String email = c.getString("email");
                    String address = c.getString("address");
                    String gender = c.getString("gender");

                    // Phone node is JSON Object
                    JSONObject phone = c.getJSONObject("phone");
                    String mobile = phone.getString("mobile");
                    String home = phone.getString("home");
                    String office = phone.getString("office");

                    // tmp hash map for single contact
                    HashMap<String, String> contact = new HashMap<>();

                    // adding each child node to HashMap key => value
                    contact.put("id", id);
                    contact.put("name", name);
                    contact.put("email", email);
                    contact.put("mobile", mobile);

                    // adding contact to contact list
                    contactList.add(contact);
                }
            } catch (final JSONException e) {
                Log.e(TAG, "Json parsing error: " + e.getMessage());
                runOnUiThread(new Runnable() {
                    @Override
                    public void run() {
                        Toast.makeText(getApplicationContext(),
                                "Json parsing error: " + e.getMessage(),
                                Toast.LENGTH_LONG)
                                .show();
                    }
                });

            }
        } else {
            Log.e(TAG, "Couldn't get json from server.");
            runOnUiThread(new Runnable() {
                @Override
                public void run() {
                    Toast.makeText(getApplicationContext(),
                            "Couldn't get json from server. Check LogCat for possible errors!",
                            Toast.LENGTH_LONG)
                            .show();
                }
            });

        }

        return null;
    }

    @Override
    protected void onPostExecute(Void result) {
        super.onPostExecute(result);
        // Dismiss the progress dialog
        if (pDialog.isShowing())
            pDialog.dismiss();
        /**
         * Updating parsed JSON data into ListView
         * */
        ListAdapter adapter = new SimpleAdapter(
                MainActivity.this, contactList,
                R.layout.list_item, new String[]{"name", "email",
                "mobile"}, new int[]{R.id.name,
                R.id.email, R.id.mobile});

        lv.setAdapter(adapter);
    }

}

}

这是我发现互联网的代码。它运作良好，但我的问题是它正在获取具有数组名称的数组。这是Json网址：http://api.androidhive.info/contacts/

这个应用程序还有另一个名为HttpHandler的类：

package info.androidhive.jsonparsing;

import android.util.Log;

import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.ProtocolException;
import java.net.URL;

public class HttpHandler {

private static final String TAG = HttpHandler.class.getSimpleName();

public HttpHandler() {
}

public String makeServiceCall(String reqUrl) {
    String response = null;
    try {
        URL url = new URL(reqUrl);
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        conn.setRequestMethod("GET");
        // read the response
        InputStream in = new BufferedInputStream(conn.getInputStream());
        response = convertStreamToString(in);
    } catch (MalformedURLException e) {
        Log.e(TAG, "MalformedURLException: " + e.getMessage());
    } catch (ProtocolException e) {
        Log.e(TAG, "ProtocolException: " + e.getMessage());
    } catch (IOException e) {
        Log.e(TAG, "IOException: " + e.getMessage());
    } catch (Exception e) {
        Log.e(TAG, "Exception: " + e.getMessage());
    }
    return response;
}

private String convertStreamToString(InputStream is) {
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();

    String line;
    try {
        while ((line = reader.readLine()) != null) {
            sb.append(line).append('\n');
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return sb.toString();
}

}

我的问题是如何将此代码编辑到另一个网址。我的json网址在这里：http://mysafeinfo.com/api/data?list=englishmonarchs&format=json

如您所见，没有数组名称。我该如何处理？欢迎任何帮助

Answer 1

使用

JsonArray listItem = new JsonArray(responseString);

无需姓名。

Answer 2

你必须直接获得JSONArray对象非常简单：

`

@Override
protected Void doInBackground(Void... arg0) {
    HttpHandler sh = new HttpHandler();

    // Making a request to url and getting response
    String jsonStr = sh.makeServiceCall(url);

    Log.e(TAG, "Response from url: " + jsonStr);

    if (jsonStr != null) {
        try {
            // Getting JSON Array node
            JSONArray contacts = new JSONArray(jsonStr);

            // looping through All Contacts
            for (int i = 0; i < contacts.length(); i++) {
                JSONObject c = contacts.getJSONObject(i);

                String id = c.getString("nm");
                String name = c.getString("cty");
                String email = c.getString("hse");
                String address = c.getString("yrs");


                // tmp hash map for single contact
                HashMap<String, String> contact = new HashMap<>();

                // adding each child node to HashMap key => value
                contact.put("id", id);
                contact.put("name", name);
                contact.put("email", email);
                contact.put("address", address);

                // adding contact to contact list
                contactList.add(contact);
            }
        } catch (final JSONException e) {
            Log.e(TAG, "Json parsing error: " + e.getMessage());
            runOnUiThread(new Runnable() {
                @Override
                public void run() {
                    Toast.makeText(getApplicationContext(),
                            "Json parsing error: " + e.getMessage(),
                            Toast.LENGTH_LONG)
                            .show();
                }
            });

        }
    } else {
        Log.e(TAG, "Couldn't get json from server.");
        runOnUiThread(new Runnable() {
            @Override
            public void run() {
                Toast.makeText(getApplicationContext(),
                        "Couldn't get json from server. Check LogCat for possible errors!",
                        Toast.LENGTH_LONG)
                        .show();
            }
        });

    }

    return null;
}

`  试试这个......希望它对你有用。

Answer 3

还有一个简单的解决方案。您可以使用谷歌Gson库来解析json。您所要做的就是根据json格式创建一个类i-e变量名与json string / file中json对象的名称相同。