从 Android 上的 URL 简单解析 JSON 并显示在列表视图中
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【中文标题】从 Android 上的 URL 简单解析 JSON 并显示在列表视图中【英文标题】：Simple parse JSON from URL on Android and display in listview 【发布时间】：2012-10-23 04:06:41 【问题描述】：我正在尝试解析从我的 android 应用中的 URL 获取的 JSON 结果...
我在 Internet 上尝试了一些示例，但无法正常工作。 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
【问题讨论】：
【参考方案1】：您可以使用AsyncTask，您必须进行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：
onPreExecute() - 最常用于设置和启动进度对话框
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）。
onPostExecute() - 这里我们脱离了后台线程，所以我们可以对响应数据进行用户界面操作，或者简单地将响应分配给特定的变量类型。
首先我们将启动该类，初始化一个String 以将结果保存在方法之外但在类内部，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);
);
然后我们需要设置连接以及我们希望如何处理响应：
@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，在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>
【讨论】：
BufferedReader 实例应该是这样的：new BufferedReader(new InputStreamReader(inputStream, "utf-8"), 8)。 JSON的默认编码是UTF，为什么要转换成另一种非全局编码？
@metallica 你是对的，我在我的一个应用程序中使用了这段代码，我没有阅读 JSON，我没有听懂。谢谢。
HttpClient 现已弃用，请使用 HttpURLConnection。
'ctrl+Q' on 'AsyncTask' 会给你很好的解释和例子
旧的正确答案，Volley Library 会为你做这一切，不再需要关心异步任务。 ref【参考方案2】：
我建议使用JSONParser 类。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，创建此类的一个实例。如果需要，您可能需要传递构造函数“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();
【讨论】：
这些变量不应该是静态的。同时使用这个类的多个实例会变得非常混乱。应该是私人的。也不要使用 == 来比较字符串：***.com/questions/513832/…【参考方案3】：JSONObject(html).getString("name");
如何获取html 字符串：
Make an HTTP request with android
【讨论】：
嘿...我收到此错误：puu.sh/1mF6v .. 当我使用代码时，这是什么意思？ .. 我在所有示例中都有错误。 该错误表示您正在 UI 线程上运行网络操作，这在早期的 Android 版本中是不受欢迎的；从 HoneyComb 开始，它会抛出一个RunTimeException 。 Read this.【参考方案4】：
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
import java.util.HashMap;
public class GetJsonFromUrl
String url = null;
public GetJsonFromUrl(String url)
this.url = url;
public String GetJsonData()
try
URL Url = new URL(url);
HttpURLConnection connection = (HttpURLConnection) Url.openConnection();
InputStream is = connection.getInputStream();
BufferedReader br = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line;
while ((line = br.readLine()) != null)
sb.append(line);
line = sb.toString();
connection.disconnect();
is.close();
sb.delete(0, sb.length());
return line;
catch (Exception e)
return null;
这个类用于发布数据
import android.util.Log;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.DataOutputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.UnsupportedEncodingException;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLEncoder;
import java.util.HashMap;
import java.util.Map;
import javax.net.ssl.HttpsURLConnection;
/**
* Created by user on 11/2/16.
*/
public class sendDataToServer
public String postdata(String requestURL,HashMap<String,String> postDataParams)
try
String response = "";
URL url = new URL(requestURL);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setReadTimeout(15000);
conn.setConnectTimeout(15000);
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter(
new OutputStreamWriter(os, "UTF-8"));
writer.write(getPostDataString(postDataParams));
writer.flush();
writer.close();
os.close();
String line;
BufferedReader br=new BufferedReader(new InputStreamReader(conn.getInputStream()));
while ((line=br.readLine()) != null)
response+=line;
Log.d("test", response);
return response;
catch (Exception e)
return e.toString();
public String postjson(String url,String json)
try
URL obj = new URL(url);
HttpURLConnection con= (HttpURLConnection) obj.openConnection();
//add reuqest header
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
String urlParameters = ""+json;
// Send post request
con.setDoOutput(true);
con.setDoInput(true);
con.setRequestProperty("Content-Type", "application/json");
OutputStreamWriter wr = new OutputStreamWriter(con.getOutputStream());
wr.write(urlParameters);
wr.flush();
wr.close();
int responseCode = con.getResponseCode();
System.out.println("\nSending 'POST' request to URL : " + url);
System.out.println("Post parameters : " + urlParameters);
System.out.println("Response Code : " + responseCode);
BufferedReader in = new BufferedReader(
new InputStreamReader(con.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null)
response.append(inputLine);
in.close();
//print result
System.out.println(response.toString());
return response.toString();
catch(Exception e)
return e.toString();
private String getPostDataString(HashMap<String, String> params) throws UnsupportedEncodingException
StringBuilder result = new StringBuilder();
boolean first = true;
for(Map.Entry<String, String> entry : params.entrySet())
if (first)
first = false;
else
result.append("&");
result.append(URLEncoder.encode(entry.getKey(), "UTF-8"));
result.append("=");
result.append(URLEncoder.encode(entry.getValue(), "UTF-8"));
return result.toString();
/* public String postdata(String url)
*/
【讨论】：
【参考方案5】：尝试：
// your get json request to server..
HttpResponse response = httpClient.execute(httpPost);
HttpEntity entity = response.getEntity();
if(entity != null)
JSONObject respObject = new JSONObject(EntityUtils.toString(entity));
String active = respObject.getString("active");
String name = respObject.getString("name");
String tab1_text = respObject.getString("tab1_text");
//....
else
//Do something here...
请参阅此示例以获取和解析来自服务器的 json 响应：
http://adblogcat.com/parse-json-data-from-a-web-server-and-display-on-listview/
【讨论】：
嗨，你知道怎么解决***.com/questions/34591855/store-php-value-to-java吗？【参考方案6】：HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet();
request.setURI(new URI(url));
HttpResponse response = client.execute(request);
BufferedReader in = new BufferedReader(new InputStreamReader(response
.getEntity().getContent()));
String line = "";
while ((line = in.readLine()) != null)
JSONObject jObject = new JSONObject(line);
if (jObject.has("name"))
String temp = jObject.getString("name");
Log.e("name",temp);
【讨论】：
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