[Solved] Set value for Spinner with custom Adapter in Android

@Haresh Chhelana example is good, However if you want to show both name and code in spinner after selecting, check this out. List<Map<String, String>> items = new ArrayList<Map<String, String>>(); for (int i = 0; i < JA.length(); i++) { json = JA.getJSONObject(i); mapData = new HashMap<String, String>(); mapData.put(“name”, json.getString(“Name”)); mapData.put(“code”, json.getString(“Code”)); items.add(mapData); } SimpleAdapter adapter … Read more

[Solved] ActivityNotFoundException: Unable to find explicit activity class {co.edu.unimagdalena.projecto/co.edu.unimagdalena.projecto.informacion2}

ActivityNotFoundException: Unable to find explicit activity class {co.edu.unimagdalena.projecto/co.edu.unimagdalena.projecto.informacion2} solved ActivityNotFoundException: Unable to find explicit activity class {co.edu.unimagdalena.projecto/co.edu.unimagdalena.projecto.informacion2}

[Solved] card view shadow effect only top side react native [closed]

install this librery npm i react-native-drop-shadow import this line import DropShadow from “react-native-drop-shadow”; use this view ginv to shasow <DropShadow style={{ shadowColor: “#000”, shadowOffset: { width: 0, height: 0, }, shadowOpacity: 1, shadowRadius: 5, }} > —– Your Contain ————– </DropShadow> solved card view shadow effect only top side react native [closed]

[Solved] Is the code after ‘break’ executed? [closed]

As per the specifications 6.6.1 The break statement [stmt.break] The break statement shall occur only in an iteration-statement or a switch statement and causes termination of the smallest enclosing iteration-statement or switch statement; control passes to the statement following the terminated statement, if any. Hence 1 should not even reach . Some Java compiler might … Read more

[Solved] PHP __get magic method unexpected behaviour

The problem is that the __get method is not reentrant, so when you access $reader->cost it is called for the first time and will call Reader::getCost(), but then $this->cost forces PHP to do a recursive call to __get which will be denied. I don’t know if it’s a bug or a feature. Read this page … Read more

[Solved] TypeError: ‘NoneType’ object has no attribute ‘__getitem__’. Bing search [closed]

Your function doesn’t return anything. Printing is not the same thing as returning a value. The default return value for a function is None, which causes your error. You want to return something, most likely r.json(): def request(query, **params): query = (‘%27’+query+ ‘%27’) r = requests.get(URL % {‘query’: query}, auth=(”, API_KEY)) return r.json() then loop … Read more

[Solved] What is the difference between Visual C++ 2010 and the Express version? [closed]

Wikipedia tells us: Visual C++ Express IDE does not have out-of-box support for compiling 64-bit applications. An x64 cross-compiler (Cl.exe) is supplied with the full version Windows SDK (available free of charge, as a separate download). Integration of 64-bit compilers to the Visual C++ 2008 Express is possible, but remains cumbersome. Visual C++ Express 2010 … Read more

[Solved] Error in My PHP [closed]

On Line 4 and 8 you need to add an opening brace change line 4 if ($hash->validate($_GET[‘serial’])) to if ($hash->validate($_GET[‘serial’])) { change line 8 if (!$_handlerLoginServer->select($_queryLogin[‘checkIP’],array($_SERVER[‘REMOTE_ADDR’]))) to if (!$_handlerLoginServer->select($_queryLogin[‘checkIP’],array($_SERVER[‘REMOTE_ADDR’]))) { 1 solved Error in My PHP [closed]

[Solved] Create a Array from HashMap in Java

Try this HashMap<String, String> param = new HashMap<String, String>(); param.put(“A”, “2”); param.put(“B”, “3”); param.put(“C”, “2”); String[] list = new String[7];// It is better to use List other than an array int i = 0; for (Map.Entry<String, String> entry : param.entrySet()) { int lim=Integer.parseInt(entry.getValue()); for(int j=0;j<lim;j++){ list[i]=entry.getKey()+” “+String.valueOf(j+1); i++; } } If you really want to … Read more

[Solved] Javascript reference [closed]

y = x; y = 6; Here, you’re assigning the value of x to y and then changing y. This doesn’t change x because x is never assigned a new value. var x = {}, y = {}; Here, you create two new objects and assign their addresses to x and y. The two variables … Read more