【发布时间】:2016-11-24 06:11:20
【问题描述】:
我正在尝试在父列表 (MainActivity) 中创建一个只有“名称”的列表视图,并将剩余的“电子邮件”和“移动设备”与“名称”一起传递给另一个活动 (SingleContactActivity)。我怎样才能做到这一点?现在我的 MainActivity 列表视图显示了所有详细信息、姓名、电子邮件、手机,我知道我必须从这里删除一些东西。我的第二个活动结果还可以。但 MainActivity 必须仅显示带有“名称”的列表。为此,我需要更改 MainActivity 代码。问题是我是 JAVA 和 Android 编程的新手。我希望您能准确显示要从 MainActivity 中删除哪一行,以便它在列表中仅显示“名称”,并将其他两个参数“电子邮件”和“移动设备”与“名称”一起传递给第二个活动。
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);
}
}
}
【问题讨论】:
-
使用意图传递数据。
-
我看不到任何启动第二个活动的代码。无论如何,您可以在意图包中传递您的数据。
-
创建一个自定义适配器并将列表视图中的方法公开给调用活动
-
没有点击处理程序,也没有意图创建。
标签: android json listview android-intent