how do you chop the last 7 digits off of a String in JAVA?

[peter moss]

Hi,
I have a string that looks something like this
String test = "test0001234567";
I want to just take the last 7 digits of that string so I can do stuff with just those 7 digits later (i.e. put them into a database)
How do I do that?
Thanks

[Rob Ross]

This is definately a Java-In-General question so I'm moving it there.
You want to use the substring method of class String, btw. Check out the JavaDocs for String.substring.

[Joe Pluta]

Just be careful - the parameters for the substring method were definitely designed by compiler guys .
The first parameter is cool - it's the index of the first character. But the send parameter is NOT the number of characters. It is also not the index of the last character in the substring. It is the index of the last character PLUS ONE. That's cool if you need to use that index as the beginning of your next call to substring, but besides that it's VERY confusing.
Luckily for you, there is a version of substring that doesn't require the second parameter. Using that one is a little easier. If there are N character in the string, you need to start at index N-7. That will get you the last seven characters of the string.
String last7 = test.substring(test.length()-7);
Note that this will blow up if the string is less than seven characters long.
Joe

[peter moss]

Thanks. That was very helpful..
I have another question though. Say I now have a bunch of ID numbers in a string called
idnumbers[]
the values lets say are 1234567, 7654321, 9999967, 2222222
Now how do I just disiplay the ones that END in 67 ?
Thanks a ton.

[Jason Menard]

Use the endsWith() method of String. For example:

String s = "1234567";
if (s.endsWith("67)) {
    System.out.println("s ends in 67");
}

[ August 29, 2003: Message edited by: Jason Menard ]