[Solved] Java equals function not working correctly [closed]

may be case is not matchingyou can try 1) if (newString.matches(aVar)){ } or 2) if (newString.equalsIgnoreCase(aVar)){ } or try with newString.trim() then compare trim() method removes spaces after and before the variable like if String s=”abc “; //space at last s.trim() will remove last space and return “abc” 3 solved Java equals function not working … Read more

[Solved] save text message in Shared Preferences [duplicate]

I don’t see why you would like to do that, but sure. If saving to SharedPrefs is all you want, it’s quite straight forward SharedPreferences sharedPref = getPreferences(Context.MODE_PRIVATE); SharedPreferences.Editor editor = sharedPref.edit(); editor.putString(“yourMessageKey”, message); editor.commit(); And of course, to read it back you SharedPreferences sharedPref = getPreferences(Context.MODE_PRIVATE); String message = sharedPref.getString(“yourMessageKey”, defaultValue); 2 solved save … Read more

[Solved] How to split String from one word to another?

This code should work. String info = “You have 2$ on your public transport card and one active ticket which expires on 2017-08-09 23.59”; Pattern pattern = Pattern.compile(“(\\d\\$).*and\\s(.*)”); Matcher m = pattern.matcher(info); while (m.find()) { System.out.println(“First Group: ” + m.group(1) + ” \nSecond Group: ” + m.group(2)); } Just like Andreas said before, you should … Read more

[Solved] How to parse a JSON string in Android?

Try using following code : try { JSONArray jsonArray = new JSONArray(response); JSONObject jsonObject = jsonArray.get(0); JSONObject _id = jsonObject.getJSONObject(“_id”); String old = _id.getString(“$oid”); String member_id = jsonObject.getString(“member_id”); String sensor_name = jsonObject.getString(“sensor_name”); String value = jsonObject.getString(“value”); String date = jsonObject.getString(“date”); }catch(JSONException e){ } 1 solved How to parse a JSON string in Android?

[Solved] force close android app in devices with android 4

Answer link : NoClassDefFoundError below SDK 21 I resolved the issue by adding this to my Application Class. @Override protected void attachBaseContext(Context base) { super.attachBaseContext(base); MultiDex.install(this); } If you don’t have/use Application class, you can put this: android:name=”android.support.multidex.MultiDexApplication” Into your tag on AndroidManifest.xml If you already have implemented an Application class Also obviously, you need … Read more

[Solved] Developing android apps [closed]

You would need a good understanding of Android fundamentals, XML layouts and SQLite db. Android training is a great place to start. A good sample app to start with all the DB stuffs comes with the SDK installation. Path is: \samples\android-17\NotePad 6 solved Developing android apps [closed]

[Solved] How to construct a button on runtime (Android) [closed]

you can create button from code. you can find all the alternative methods of setting attribute in the doc. for example Button button = new Button(this); button.setText(“hi”); button.setId(Id); button.setTextColor(Color.Red); button.setBackgroundResource(R.drawable.icon); buttonlayout.addView(button); 1 solved How to construct a button on runtime (Android) [closed]