我正在尝试打开我的Android应用程序活动,点击在短信中收到的链接,这是我的活动代码打开。在这个应用程序中,我正在使用Http服务。
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main2);
Bundle bundle = getIntent().getExtras();
user = bundle.getString("Username");
pass = bundle.getString("Password");
//initializing listview and hero list
listView = (ListView) findViewById(R.id.listView);
tktList = new ArrayList<>();
//this method will fetch and parse the data
loadHeroList();
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
MenuInflater inflater = getMenuInflater();
inflater.inflate(R.menu.main_menu, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
switch (item.getItemId()) {
case R.id.logout:
SharedPreferences SM = getSharedPreferences("userrecord", 0);
SharedPreferences.Editor edit = SM.edit();
edit.putBoolean("username", false);
edit.commit();
Intent intent = new Intent(Main2Activity.this, MainActivity.class);
startActivity(intent);
finish();
}
switch (item.getItemId()){
case R.id.user:
Toast.makeText(getApplicationContext(), user, Toast.LENGTH_SHORT).show();
}
return super.onOptionsItemSelected(item);
}
private void loadHeroList() {
//getting the progressbar
final ProgressBar progressBar = (ProgressBar) findViewById(R.id.progressBar);
//making the progressbar visible
progressBar.setVisibility(View.VISIBLE);
//creating a string request to send request to the url
StringRequest stringRequest = new StringRequest(Request.Method.GET, "http://182.18.163.39/list_details.php?username="+user+"&key="+pass,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
//hiding the progressbar after completion
progressBar.setVisibility(View.INVISIBLE);
try {
JSONArray jsonarray = new JSONArray(response);
for (int i = 0; i < jsonarray.length(); i++) {
JSONObject jsonobject = jsonarray.getJSONObject(i);
String name = jsonobject.getString("Sno");
String Tktid = jsonobject.getString("TKTID");
link = jsonobject.getString("Link");
List list = new List(user, pass, jsonobject.getString("Sno"), jsonobject.getString("TKTID"),jsonobject.getString("Link"));
tktList.add(list);
Log.i("website content", name);
Log.i("website content", Tktid);
Log.i("website content", link);
}
//creating custom adapter object
ListViewAdapter adapter = new ListViewAdapter(tktList, getApplicationContext());
//adding the adapter to listview
listView.setAdapter(adapter);
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
//displaying the error in toast if occurrs
Toast.makeText(getApplicationContext(), error.getMessage(), Toast.LENGTH_SHORT).show();
}
});
//creating a request queue
com.android.volley.RequestQueue requestQueue = Volley.newRequestQueue(this);
//adding the string request to request queue
requestQueue.add(stringRequest);
}
}
这里我想要的是当我通过短信发送链接并点击该链接时,它将打开特定的活动文件任何人都可以帮助我完成此操作。
答案 0 :(得分:0)
您可以阅读Chrome浏览器链接,只需在清单中添加一些代码。
<activity
android:name=".MainActivity"
android:label="@string/app_name">
<intent-filter>
<action android:name="android.intent.action.VIEW"/>
<category android:name="android.intent.category.DEFAULT"/>
<category android:name="android.intent.category.BROWSABLE"/>
<data android:scheme="http"
android:host="example.com"
android:pathPrefix="/"/>
</intent-filter>
</activity>