Comparing strings by their alphabetical order

Question

String.compareTo might or might not be what you need.

Take a look at this link if you need localized ordering of strings.

Take a look at the String.compareTo method.

s1.compareTo(s2)

From the javadocs:

The result is a negative integer if this String object lexicographically precedes the argument string. The result is a positive integer if this String object lexicographically follows the argument string. The result is zero if the strings are equal; compareTo returns 0 exactly when the equals(Object) method would return true.

String a = "..."; 
String b = "...";  

int compare = a.compareTo(b);  

if (compare < 0) {  
    //a is smaller
}
else if (compare > 0) {
    //a is larger 
}
else {  
    //a is equal to b
} 

You can call either string's compareTo method (java.lang.String.compareTo). This feature is well documented on the java documentation site.

Here is a short program that demonstrates it:

class StringCompareExample {
    public static void main(String args[]){
        String s1 = "Project"; String s2 = "Sunject";
        verboseCompare(s1, s2);
        verboseCompare(s2, s1);
        verboseCompare(s1, s1);
    }

    public static void verboseCompare(String s1, String s2){
        System.out.println("Comparing \"" + s1 + "\" to \"" + s2 + "\"...");

        int comparisonResult = s1.compareTo(s2);
        System.out.println("The result of the comparison was " + comparisonResult);

        System.out.print("This means that \"" + s1 + "\" ");
        if(comparisonResult < 0){
            System.out.println("lexicographically precedes \"" + s2 + "\".");
        }else if(comparisonResult > 0){
            System.out.println("lexicographically follows \"" + s2 + "\".");
        }else{
            System.out.println("equals \"" + s2 + "\".");
        }
        System.out.println();
    }
}

Here is a live demonstration that shows it works: http://ideone.com/Drikp3