Java Converting String into Integer for Comparison

Question

I am writing a compareTo method for the Comparable Interface for a specific class. I need to take two different strings and compare them and see if one holds a greater value than the other. I would normally use .compareTo() but in this case it is giving me an error and not working. This is what I have:

public int compareTo(Object o) {
    if (name < ((AccountOwnerComparable) o).name)
        return -1;
    else if (name > ((AccountOwnerComparable) o).name)
        return 1;
    else
        return 0;

}

I know I can't use the < or > but it is just to show what the scenario is.

Nishant Shah · Accepted Answer · 2012-12-02 18:16:12Z

1

You might have missed a bracket or something while casting. Try the following piece of code.

public int compareTo(Object o) {
    if (name.compareTo(((AccountOwnerComparable) o).name) > 0) {
        System.out.println("Greater");
        return -1;
    } else if (name.compareTo(((AccountOwnerComparable) o).name) < 0) {
        System.out.println("Lesser");
        return 1;
    } else {
        System.out.println("Equal");
        return 0;
    }

}

answered Dec 2, 2012 at 18:16

Nishant Shah

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Comments

arshajii · Accepted Answer · 2012-12-02 18:14:54Z

1

The Comparable interface is generic. Your class declaration should probably look something like this:

class AccountOwnerComparable implements Comparable<AccountOwnerComparable> {...

Notice the AccountOwnerComparable type parameter. With this, you would have

public int compareTo(AccountOwnerComparable other) {
    return name.compareTo(other.name);
}

answered Dec 2, 2012 at 18:14

arshajii

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Comments

dreamcrash · Accepted Answer · 2012-12-02 18:45:15Z

You just have to do:

name.compareTo(String_to_be_compare);

compareTo

public int compareTo(Object o)

Compares this String to another Object. If the Object is a String, this function behaves like compareTo(String).

Note that:

Returns:

The value 0 if the argument is a string lexicographically equal to this string; a value less than 0 if the argument is a string lexicographically greater than this string; and a value greater than 0 if the argument is a string lexicographically less than this string. (source)

So to your case:

public int compareTo(Object o)
{
    AccountOwnerComparable obj = (AccountOwnerComparable) o;
    return obj.name.compareTo(this.name);
}

tw: implement the comparable interface specifying what you want to compare to (usually the same type), since you did not specify anything your interface compares to anything which is odd.

andreih · Accepted Answer · 2012-12-02 17:41:32Z

0

Use Integer.parseInt() method. In this way you could use > or < and it is more verbose. I assume you want to compare two String objects containing numbers. Although you did not write it in the questions, the title says so.

answered Dec 2, 2012 at 17:41

andreih

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2 Comments

user1729448 Over a year ago

They are not Integers/. Both are Strings just containing letters. The compareTo() method returns an integer when it compares two different Strings. That is what I need

andreih Over a year ago

well, the title was misleading.

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