我在php中显示名称,emial id,ph no no数据库到列表视图。
enter code here
public class MainActivity extends ListActivity
{
// url to make request
private static String url = "http://10.0.2.2/programs/get_data.php";
private static final String TAG_CONTACTS = "contacts";
private static final String TAG_ID = "id";
private static final String TAG_NAME = "name";
private static final String TAG_phno = "phno";
JSONArray contacts;
@Override
public void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
//setContentView(R.layout.activity_main);
// Hashmap for ListView
ArrayList<HashMap<String, String>> contactList = ArrayList<HashMap<String,string();
// Creating JSON Parser instance
JSONParser jParser = new JSONParser();
// getting JSON string from URL
JSONObject json = jParser.getJSONFromUrl(url);
try {
// Getting Array of Contacts
contacts = json.getJSONArray(TAG_CONTACTS);
// looping through All Contacts
for(int i = 0; i < contacts.length(); i++)
{
JSONObject c = contacts.getJSONObject(i);
String name = c.getString("name");
String phno = c.getString("phno");
String email = c.getString("id");
System.out.println("hi");
// creating new HashMap
HashMap<String, String> map = new HashMap<String, String>();
map.put(TAG_NAME, name);
map.put(TAG_ID,email);
map.put(TAG_phno, phno);
// adding HashList to ArrayList
contactList.add(map);
}
}catch (JSONException e) {
e.printStackTrace();
}
ListAdapter adapter = new SimpleAdapter(this, contactList,
R.layout.list_item,
new String[] { TAG_NAME, TAG_ID, TAG_phno }, new int[] {
R.id.name, R.id.email, R.id.phno });
setListAdapter(adapter);
}
}
public class JSONParser extends Activity
{
static InputStream is;
static JSONObject jObj;
static String json = "";
// constructor
public JSONParser()
{
}
public JSONObject getJSONFromUrl(String url)
{
// Making HTTP request
try {
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;
}
}
我收到错误
11-21 22:44:00.306:E / JSON Parser(17671):解析数据时出错org.json.JSONException:值[{“id”:“c.com”,“phno”:“12”, “name”:“Chaitra D”},{“id”:“s.com”,“phno”:“1254”,“name”:“Shrivatsa K”},{“id”:“m.com”, “phno”:“1456”,“name”:“Mahalaxmi B”},{“id”:“Sh.com”,“phno”:“1234”,“name”:“Sharadha”},{“id” :“shh.com”,“phno”:“1564”,“name”:“shrilatha”},{“id”:“n.com”,“phno”:“97”,“name”:“namratha h “},{”id“:”shri.com“,”phno“:”167“,”name“:”Shrinivas D“},{”id“:”p.com“,”phno“:”987“ ,“name”:org.json.JSONArray类型的“PG sunita”}]无法转换为JSONObject
请帮助我谢谢
答案 0 :(得分:0)
由于你没有json对象,你得到了这个异常,你有一个json对象的json数组。首先,将您的字符串解析为json数组,然后循环遍历数组以获取每个对象。
答案 1 :(得分:0)
我猜服务器的json是一个数组,所以你应该从该数组获得第一个JSONObject。 就像:
public JSONArray getJSONFromUrl(String url) {
// get JSONArray... }
jsonArray = getJSONFromUrl(Url); JSONOBject jsonObject = jsonArray.getJSONObject(0);
或者更简单的方法是使用此lib从服务器获取JSON:http://loopj.com/android-async-http/
AsyncHttpClient client = new AsyncHttpClient();
client.get(stringURL, paramsOrNull, new JsonHttpResponseHandler() {
@Override
public void onStart() {
// do somethings when start send request to server
}
@Override
public void onFailure(Throwable arg0) {
// do somethings when fail to request or receive respone from server
}
@Override
public void onSuccess(JSONObject rawJson) {
// do somethings when sucessful...
}