Testing string and integer arrays against user input

Question

I have created arrays for strings and integers I want to use in my program and I want to use them instead of using

(name.equals "barry"||"matty"

for example.

I want to know how to write the if statement to check the user input against the strings in the array.

import java.util.Scanner;

public class Username
{
    public static void main (String[]args)
    {
        Scanner kb = new Scanner (System.in);
        // array containing usernames 
        String [] name = {"barry", "matty", "olly","joey"}; 

        System.out.println("Enter your name");
        name = kb.nextLine();
        if (name.equals("barry ")|| name.equals("matty" ) || name.equals("olly")||name.equals("joey"))
            System.out.println("you are verified you may use the lift");

        Scanner f = new Scanner(System.in);
        int floor;

        int [] floor = {0,1,2,3,4,5,6,7};

        System.out.println("What floor do you want to go to ");
        floor = f.nextInt();

        if (floor >7)
            System.out.println("Invalid entry");
        else if (floor <= 7)
            System.out.println("Entry valid");
    }
}

Jon Skeet · Accepted Answer · 2014-03-29 13:50:30Z

I think you're just looking for List.contains - but that requires a List rather than an array, of course. There are two obvious options here.

Firstly, you could use a List<String> to start with:

List<String> names = new ArrayList<>();
names.add("barry");
names.add("matty");
names.add("olly");
names.add("joey");
...
if (names.contains(name))
{
    ...
}

Alternatively, you could use Arrays.asList to create a view:

String[] names = {"barry", "matty", "olly", "joey"}; 
List<String> namesList = Arrays.asList(names);
...
if (namesList.contains(name))
{
}

As a third option, if you put make your names array sorted (either by hand or by calling Arrays.sort) you could use Arrays.binarySearch to try to find the name entered by the user:

String[] names = {"barry", "matty", "olly", "joey"}; 
Arrays.sort(names);
...
if (Arrays.binarySearch(names, name) >= 0)
{
    ...
}

JB Nizet · Accepted Answer · 2014-03-29 13:50:59Z

2

Arrays are very low-level structures that don't provide any method. You'd better use collections instead, which have a contains() method:

Set<String> names = new HashSet<>(Arrays.asList(new String[] {"barry", "matty", "olly","joey"}));

if (names.contains(name)) {
    ...
}

Since you don't seem to care about the order of the names, but only want to test if the collection contains a name or not, a HashSet is the best data structure: HashSet.contains() runs in constant time (O(1)), whereas List.contains(), for example, is O(n).

Read the collections tutorial.

answered Mar 29, 2014 at 13:50

JB Nizet

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Comments

Salah · Accepted Answer · 2014-03-29 13:59:48Z

2

You can loop through your array instead:

String name = kb.nextLine();
if(contains(name)) {
    System.out.println("you are verified you may use the lift");
}


public boolean contains(String name) {
    String [] names = {"barry", "matty", "olly","joey"}; 
    for (int i = 0; i < names.length; i++) {
        if(names[i].equals(name)) {
            System.out.println("you are verified you may use the lift");
        }
    }

Or you can use List.contains(), in this case you have to add your names inside List instead of regular array.

For Example:

String[] names = {"barry", "matty", "olly", "joey"}; 
List<String> namesList= Arrays.asList(names);

if (namesList.contains(name)) {
    System.out.println("you are verified you may use the lift");
}

edited Mar 29, 2014 at 13:59

answered Mar 29, 2014 at 13:48

Salah

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Comments

AndreaTaroni86 · Accepted Answer · 2014-03-29 13:54:49Z

1

Using ArrayList and List instead of Array of Strings:

List<String>names = new ArrayList<String>(names); 
name = kb.nextLine();
if(names.indexOf(name)>-1)System.out.println("you are verified you may use the lift");

This because of the indexof in List returns -1 if not found or the index of the founded element, starting with 0.

I think that also

if(names.contains(name))System.out.println("you are verified you may use the lift");

works

answered Mar 29, 2014 at 13:54

AndreaTaroni86

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Comments

Martin Prikryl · Accepted Answer · 2014-03-29 14:11:08Z

0

Use a for loop

get input
for int i = 0, i < array size i++
if input.equals(array[i]) then do stuff

If you switch to an arraylist, you can do if(array.contains(input))

edited Mar 29, 2014 at 14:11

Martin Prikryl

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answered Mar 29, 2014 at 13:52

raranak

11

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