You can send you HTML as a JSON object. Here a quick example. var elm = document.getElementById(‘yourElement’); var value = elm.innerHTML; var objToSend = JSON.stringify(value); // now send using ajax … 1 solved Sending HTML to backend
import java.io.*; import java.util.ArrayList; import java.util.*; public class ArrayOperations { public static void main(String[] args) throws Exception{ List<Integer> readFromFile = readFromFile(“data.txt”) ; printArray(readFromFile) ; } public static List<Integer> readFromFile(String fileName) throws FileNotFoundException { File f = new File(fileName); Scanner fileIn = new Scanner (f); List<Integer> list = new ArrayList<Integer>(); while (fileIn.hasNextInt()){ // quit when you … Read more
if(temp[i].equals(object[j])) // It shows arrayout of bound exception Don’t want personally to be “Captain Obvious”, but it’s a good idea to check an index vs the array length before trying to access array item by index. For example: if((object.length > j) && (temp[i].equals(object[j]))) And yes, use “Array.equals()” to compare arrays, as @Prateek proposed before. solved … Read more
Java is the standard way of writing Android apps, but it’s not strictly necessary. For example, there’s also Xamarin.Android which lets you write Android apps in C# – although it will still fire up a Dalvik VM behind the scenes, as the Android “native” controls are in Java. Using Java is probably the simplest option. … Read more
ArrayList is a list implementation that’s backed by an Object[]. It supports random access and dynamic resizing. LinkedList is a list implementation that uses references to head and tail to navigate it. It has no random access capabilities, but it too supports dynamic resizing. Bear in mind that both support the get(int index) signature, but … Read more
Use orElseGet: Optional<String> startingOptional = getMyOptional(); String finishingValue = startingOptional.orElseGet(() -> { System.out.println(“value not found”); return “”; }); Using .map(value -> value) is useless: transforming a value into itself doesn’t change anything. solved Is there a way to combine Java8 Optional returning a value with printing a message on null?
You need to update the left and right values not the middle..change it like this while (true) { middle = (left + right) / 2; int copmarison = Name.compareTo(StudentName[middle]); if (Name.equals(StudentName[middle])) { return middle; } else if (left > right) { return count; } else { if (copmarison > 0) { left = middle + … Read more
You might try: ^(?!((\\d{9})|(\\d{3}-\\d{2}-\\d{4})|(\\d{3}-\\d{3}-\\d{3}))$).* To explain, if we read the query you provided: ^((?!\\d[9]$)|(?!(\\d{3}-?\\d{2}-?\\d{4}$)|(?!(\\d{3}-?\\d{3}-?\\d{3})$)$ We could read that: is not followed by xxxxxxxxx OR is not followed by xxx-xx-xxxx OR is not followed by xxx-xxx-xxx (in my version at the top, I rephrased this to be: is not (xxxxxxxxx OR xxx-xx-xxxx OR xxx-xxx-xxx).). Any string … Read more
In order for this to work you need to override hashcode and equals on the contents of the list. Contains uses these methods, which by default are based on the memory reference so it will only work for a shallow copy. solved .contains() on a list of objects [closed]
If you want to run some java code(In server) and get the response in your javascript(in client). You can use Ajax in jquery. http://api.jquery.com/jQuery.ajax/ Remember this is for java (advance) programming. solved how to call java object in javascript?
First you would need to create a Map of all the letters: Hashmap<String, String> map = new Hashmap<String, String>(); map.put(“a”, “c”); map.put(“b”, “f”); … To get the translation of each letter you simply get the value from the map: String translatedLetter = map.get(letter); So now you would need to create a loop to translate the … Read more
Create your Comparator: // Here T is your class name public class Test implements Comparator<T> { @Override public int compare(T o1, T o2) { return (o1.getNumber() – o2.getNumber()); } } And pass it to Collections.sort: Collections.sort(yourList, new Test()); solved How to sort an ArrayList by comparing the element’s property?
Aside from using loops explicitly, for List<Integer> list you can do: int sum = list.stream().mapToInt(Integer::intValue).sum(); If it’s an int[] arr then do: int sum = IntStream.of(arr).sum(); This is based on the use of streams. Or you can do this simple one liner loop: int sum = 0; for (Integer e : myList) sum += e; … Read more
How to place an image into ImageView from Gallery private int PICK_IMAGE_REQUEST = 1; private void openGallery() { tvGallery.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View view) { Intent intent = new Intent(); // Show only images, no videos or anything else intent.setType(“image/*”); intent.setAction(Intent.ACTION_GET_CONTENT); // Always show the chooser (if there are multiple options available) startActivityForResult(Intent.createChooser(intent, … Read more
Array variables are also references just like a class-typed variable. So x and y contain only references to an array somewhere in memory. Let’s look at that assignment: But what does happen when we use “x=y” ? Yes, just like with objects in java, the address to the array y is stored in x. The … Read more