sorting contents of strings

[Tim Tock]

hey all, having a bit of trouble, got a 3 character string, just wanna be able to remove the middle one and return a string something like
public static void main(String args[]) {
String phrase = new String ("CAT");
if (phrase.length()==3)
{
char temp = phrase.charAt(3);
char temp1= phrase.charAt(1);

String temp2 = new String ("temp");
temp1.concat(temp);
System.out.println(temp1);
}
}
Any suggestions? i've cant concat, i cant put a char straight into a string, im all out
T

[Mark Vedder]

This sounds like a homework problem, so I don't want to give the code right to you. Let me give you a couple of hints. There are a couple of different ways to approach the "problem". Keep in mind that at their heart, Stings are really char arrays (char[]). Take a look at the String API and see if with that thought in mind you might find a different way of approaching the problem. You can also take a look at the StringBuffer API for some alternatives.

[Dirk Schreckmann]

Tim,
Note that the class documentation that Mark mentioned is available at http://java.sun.com/j2se/1.4.2/docs/api/
Take a look at the documentation of the methods available to Strings. Perhaps you'll notice a few things that you'll be able to make use of.
Also, note that the index of the first character in a String is 0, not 1.
Don't hesitate to ask for further nudging in the right direction.

[Davy Kelly]

Tim,
public static void main(String args[]) {
String phrase = new String ("CAT");
if (phrase.length()==3)
{
char temp = phrase.charAt(3); //is your indexes correct?
char temp1= phrase.charAt(1);
String temp2 = new String ("temp"); //is this going to be output?
temp1.concat(temp); //is this correct
System.out.println(temp1); //what will be output here if it did work?
}
}
in your code, look at your indexes, are they correct??
is you concat() method correct?
also check the API to see if you can find a better solution. to your charAt() methods.
I am sure you can see what mistakes you have done, I was going to give the answer until i saw others thought it was a homework question
good luck, but post back with any amendments you have
Davy
[ February 28, 2004: Message edited by: Davy Kelly ]

[Tim Tock]

edited again, oops, fantastic, just got to grips with stringbuffers, heres wat i got:
public class JdbcExample2 {
public static void main(String args[]) {
String phrase = new String ("ABC");
String x;
if (phrase.length()==3)
{
char temp = phrase.charAt(0);
char temp1= phrase.charAt(2);
x = new StringBuffer().append(temp).append(temp1).toString();
System.out.println(x);
}
}
}
Thanks for the help, just have to add, I cannot thank javaranch enough for all the help and advice i've recieved since being on here (I just wish I could put something back into the board)
big THANKYOU to everyone
ps. it wasnt a homework assignmet, lol, it was just something i wanted to do for a little project of mine, but thanks all the same!!!
[ February 29, 2004: Message edited by: Tim Tock ]

[sever oon]

Here's what I'd write:

String s = "abc";
s = new StringBuffer( s ).deleteCharAt( 1 ).toString();

sev

[Tony Morris]

Stings [sic] are really char arrays (char[]).

This methodology of thinking is prone to problems.
Most VMs implement a String using a char[].
However, a char[] is a mutable type, where a String is not (intended to be anyway - this is another academic argument).
Example:
char[] a = new char[]{'a', 'b', 'c'};
a[0] = 'd';
Now let's see the String equivalent:
String a = "abc";
// What now ?
It can't be done (well it can, but like I said, that's another story).