In python, you can use a very simple if statement to see if what the user has entered (or just a variable) is in a list:
myList = ["x", "y", "z"] myVar = "x" if myVar in x: print("your variable is in the list.") How would I be able to do this in Java?
If your array type is a reference type, you can convert it to a List with Arrays.asList(T...) and check if it contains the element
if (Arrays.asList(array).contains("whatever")) // do your thing 
String[] array = {"x", "y", "z"}; if (Arrays.asList(array).contains("x")) { System.out.println("Contains!"); } Here is one solution,
String[] myArray = { "x", "y", "z" }; String myVar = "x"; if (Arrays.asList(myArray).contains(myVar)) { System.out.println("your variable is in the list."); } Output is,
your variable is in the list. You could iterate through the array and search, but I recommend using the Set Collection.
Set<String> mySet = new HashSet<String>(); mySet.add("x"); mySet.add("y"); mySet.add("z"); String myVar = "x"; if (mySet.contains(myVar)) { System.out.println("your variable is in the list"); } Set.contains() is evaluated in O(1) where traversing an array to search can take O(N) in the worst case.
Rather than calling the Arrays class, you could just iterate through the array yourself
String[] array = {"x", "y", "z"}; String myVar = "x"; for(String letter : array) { if(letter.equals(myVar) { System.out.println(myVar +" is in the list"); } } Remember that a String is nothing more than a character array in Java. That is
String word = "dog"; is actually stored as
char[] word = {"d", "o", "g"}; So if you would call if(letter == myVar), it would never return true, because it is just looking at the reference id inside the JVM. That is why in the code above I used
if(letter.equals(myVar)) { } 