Sequential String search in Java [duplicate]

Question

This question already has answers here:

How do I compare strings in Java? (23 answers)

Closed 8 years ago.

Hello again stack overflow.. ers, I have created this bit of code to sequentially search through an array to find a value, in this case a name. However i cant get it to work, it wont find the name i type in, and always returns a value of -1.

When i changed all of the types to INT and did a search on the age array, it worked, so it may have something to do with the string type? Can someone please help me?

public static void main(String[] args) {

    String[] name = {"John", "bert", "Elle", "beth"};
    String[] sex = {"Male", "Male", "female", "female"};
    int[] age = {18, 25, 22, 36};
    int found;

    Scanner keyboard = new Scanner(System.in);
    System.out.println("Enter name to search: ");

    String searchName = keyboard.next();

    found = searchNames(name, searchName);

    if(found == -1)
        System.out.println("Error, not found: " + found);
    else
        System.out.println("Found At: " + found);

}

private static int searchNames(String[] name, String searchName) {

    int i = 0, foundAt = -1;
    boolean found = false;

    while (!found && i < name.length)
    {
        if (name[i] == searchName)
        {
            found = true;
            foundAt = i;
        }
        i++;
    }
    return foundAt;
}

Answer 1

This is going to be closed soon, but use .equals() to compare Strings.

Answer 2

Try replacing this portion of your code

while (!found && i < name.length)
{
    if (name[i] == searchName)
    {
        found = true;
        foundAt = i;
    }
    i++;
}

with

for(int i=0; i<name.length; i++)
{
    if(name[i].equalsIgnoreCase(searchName))
    {
        found = true;
        foundAt = i;
        break;
    }
}