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Java, cant figure out how to strip symbols from a String for a Palindrome

Im in highschool and this is an assignment i have, you guys are out of my league but im willing to learn and understand. I looked all over the place but all i could find was complicated syntax i dont know yet. This is what i have, it takes a String and reverses it. I managed to get it to ignore Capitals, but i cannot figure out how to make it ignore symbols. The numbers i have there are from the ANSI Characters, there is a list on textpad im using. Dont be afraid to be harsh, im not good at this and i only want to improve so have at it.
import java.util.Scanner;
public class PalindromeV2
{
public static void main(String[] args)
{
//declare
Scanner sc = new Scanner(System.in);
String fwd, rev;
String result;
//input
System.out.println("What word would you like to Palindrome test?");
fwd = sc.next();
rev = reverseString(fwd);
result = stripPunctuation(fwd);
if(stripPunctuation(rev).equals(stripPunctuation(fwd)))
{
System.out.println("That is a palindrome");
}
else
System.out.println("That is not a palindrome");
}
public static String reverseString(String fwd)
{
String rev = "";
for(int i = fwd.length()-1; i >= 0; i--)
{
rev += fwd.charAt(i);
}
return rev.toUpperCase();
}
public static String stripPunctuation(String fwd)
{
String result = "";
fwd = fwd.toUpperCase();
for(int i = fwd.length()-1; i >= 0; i--)
{
if((fwd.charAt(i)>=65 && fwd.charAt(i)<=90)||(fwd.charAt(i) >= 48 && fwd.charAt(i) <= 58));
result = result + fwd.charAt(i);
}
return result;
}
}

You can use this as a checking condition
if (Character.isLetter(fwd.charAt(i)) {
// do something
}
This will check to make sure the character is a letter, so you don't have to worry about case, numbers, or other symbols.

If you want to strip your string out of some set of characters than do something like that
clearString=targetStringForStripping.replaceAll([type_characters_for_stripping],"");
this will remove all characters you will provide inside square brackets.
There is even more. If you want to let say leave only letters (because in palindromes nothing matters except letters - spaces are not important to) than you simply can use predefine character set - letters.
To conclude all if you do
clearString=targetStringForStripping.replaceAll("[\w]","");
or
clearString=targetStringForStripping.replaceAll("[^a-zA-Z]","");
you will get clear string with white characters in first example, and only letters in second one. Perfect situation for isPalindrom resolution.

if((fwd.charAt(i)>=65 && fwd.charAt(i)<=90)||(fwd.charAt(i) >= 48 && fwd.charAt(i) <= 58));
you have semicolon at last. so i think if condition is no use here

Since this is a highschool assignment, I'll just give some pointers, you'll figure it out on your own.
Think about what you want to include / exclude, then write the code.
Keep in mind, that you can compare char variables using < or > operators as long as you do not want to handle complex character encodings.
A String is really just a sequence of chars which one by one you can compare or reorder, include or exclude.
A method should only do one thing, not a lot of things. Have a look at your reverseString method. This is doing an toUpperCase to your string at the same time. If your programs get more complex, this way of doing things is not to easy to follow.
Finally, if you e.g. just want to include capital letters in your palindrome check, then try some code like this:
char[] toCheck = fwd.toCharArray();
for (char c : toCheck) {
if (c >= 'A' && c <= 'Z') {
result = result + c;
}
}
Depending on your requirements this might do what you want. If you want something different, have a look at the hints I gave above.

Java golf?
public static String stripPunctuation(String stripThis) {
return stripThis.replaceAll("\\W", "");
}

Replace char at specific substring

first of all I want to say that I am kinda new to Java. So please be easy on me :)
I made this code, but I cannot find a way to change a character at a certain substring in my progress bar. What I want to do is this:
My progressbar is made out of 62 characters (including |). I want the 50th character to be changed into the letter B (uppercase).It should look something like this: |#########----B--|
I tried several things, but I dont know where to put the line of code to make this work. I tried using the substring and the replace code, but I can't find a way to make this work. Maybe I need to write my code in a different way to make this work? I hope someone can help me.
Thanks in advance!
int ecttotal = ectcourse1+ectcourse2+ectcourse3+ectcourse4+ectcourse5+ectcourse6+ectcourse7;
int ectmax = 60;
int ectavg = ectmax - ecttotal;
//Progressbar
int MAX_ROWS = 1;
for (int row = 1; row == MAX_ROWS; row++)
{
System.out.print("|");
for (int hash = 1; hash <= ecttotal; hash++)
System.out.print ("#");
for (int hyphen = 1; hyphen <= ectavg; hyphen++)
System.out.print ("-");
System.out.print("|");
}
System.out.println("");
System.out.println("");
}

Can you tell a little more what you want. Because what i sea it that, that you write some string into console. And is not way to change that what you already print to console.
Substring you can use only at String varibles.
If you want to change lettir with substring method in string varible try smth. like this:
String a="thi is long string try it";
if(a.length()>50){
a=a.substring(0,49)+"B"+a.substring(51);
}
Other way to change charater in string is to use string builder like this:
StringBuilder a= new StringBuilder("thi is long string try it");
a.setCharAt(50, 'B');
Sure you must first check the length of string to avoid the exceptions.
I hope that I helped you :)

Java StringBuilder has method setCharAt which can replace character at position with new character.
StringBuilder myName = new StringBuilder(<original string>);
myName.setCharAt(<position>, <character to replace>);
<position> starts with index 0
In your case:
StringBuilder myName = new StringBuilder("big longgggg string");
myName.setCharAt(50, 'B');

You can replace a certain index in a string by concatenating a new string around the intended index. For example the following code replaces the letter c with the letter X. Where 2 is the intended index to replace.
In other words, this code replaces the 3rd character in the string.
String s = "abcde";
s = s.substring(0, 2) + "X" + s.substring(3);
System.out.println(s);

Determining if a given string of words has words greater than 5 letters long

So, I'm in need of help on my homework assignment. Here's the question:
Write a static method, getBigWords, that gets a String parameter and returns an array whose elements are the words in the parameter that contain more than 5 letters. (A word is defined as a contiguous sequence of letters.) So, given a String like "There are 87,000,000 people in Canada", getBigWords would return an array of two elements, "people" and "Canada".
What I have so far:
public static getBigWords(String sentence)
{
String[] a = new String;
String[] split = sentence.split("\\s");
for(int i = 0; i < split.length; i++)
{
if(split[i].length => 5)
{
a.add(split[i]);
}
}
return a;
}
I don't want an answer, just a means to guide me in the right direction. I'm a novice at programming, so it's difficult for me to figure out what exactly I'm doing wrong.
EDIT:
I've now modified my method to:
public static String[] getBigWords(String sentence)
{
ArrayList<String> result = new ArrayList<String>();
String[] split = sentence.split("\\s+");
for(int i = 0; i < split.length; i++)
{
if(split[i].length() > 5)
{
if(split[i].matches("[a-zA-Z]+"))
{
result.add(split[i]);
}
}
}
return result.toArray(new String[0]);
}
It prints out the results I want, but the online software I use to turn in the assignment, still says I'm doing something wrong. More specifically, it states:
Edith de Stance states:
⇒     You might want to use: +=
⇒     You might want to use: ==
⇒     You might want to use: +
not really sure what that means....

The main problem is that you can't have an array that makes itself bigger as you add elements.
You have 2 options:
ArrayList (basically a variable-length array).
Make an array guaranteed to be bigger.
Also, some notes:
The definition of an array needs to look like:
int size = ...; // V- note the square brackets here
String[] a = new String[size];
Arrays don't have an add method, you need to keep track of the index yourself.
You're currently only splitting on spaces, so 87,000,000 will also match. You could validate the string manually to ensure it consists of only letters.
It's >=, not =>.
I believe the function needs to return an array:
public static String[] getBigWords(String sentence)
It actually needs to return something:
return result.toArray(new String[0]);
rather than
return null;
The "You might want to use" suggestions points to that you might have to process the array character by character.

First, try and print out all the elements in your split array. Remember, you do only want you look at words. So, examine if this is the case by printing out each element of the split array inside your for loop. (I'm suspecting you will get a false positive at the moment)
Also, you need to revisit your books on arrays in Java. You can not dynamically add elements to an array. So, you will need a different data structure to be able to use an add() method. An ArrayList of Strings would help you here.

split your string on bases of white space, it will return an array. You can check the length of each word by iterating on that array.
you can split string though this way myString.split("\\s+");

Try this...
public static String[] getBigWords(String sentence)
{
java.util.ArrayList<String> result = new java.util.ArrayList<String>();
String[] split = sentence.split("\\s+");
for(int i = 0; i < split.length; i++)
{
if(split[i].length() > 5)
{
if(split[i].matches("[a-zA-Z]+"))
{
result.add(split[i]);
}
if (split[i].matches("[a-zA-Z]+,"))
{
String temp = "";
for(int j = 0; j < split[i].length(); j++)
{
if((split[i].charAt(j))!=((char)','))
{
temp += split[i].charAt(j);
//System.out.print(split[i].charAt(j) + "|");
}
}
result.add(temp);
}
}
}
return result.toArray(new String[0]);
}

Whet you have done is correct but you can't you add method in array. You should set like a[position]= spilt[i]; if you want to ignore number then check by Float.isNumber() method.

Your logic is valid, but you have some syntax issues. If you are not using an IDE like Eclipse that shows you syntax errors, try commenting out lines to pinpoint which ones are syntactically incorrect. I want to also tell you that once an array is created its length cannot change. Hopefully that sets you off in the right directions.

Apart from syntax errors at String array declaration should be like new String[n]
and add method will not be there in Array hence you should use like
a[i] = split[i];
You need to add another condition along with length condition to check that the given word have all letters this can be done in 2 ways
first way is to use Character.isLetter() method and second way is create regular expression
to check string have only letter. google it for regular expression and use matcher to match like the below
Pattern pattern=Pattern.compile();
Matcher matcher=pattern.matcher();
Final point is use another counter (let say j=0) to store output values and increment this counter as and when you store string in the array.
a[j++] = split[i];

I would use a string tokenizer (string tokenizer class in java)
Iterate through each entry and if the string length is more than 4 (or whatever you need) add to the array you are returning.
You said no code, so... (This is like 5 lines of code)

How do I print content of reference in JAVA?

Newbie question comming up. Trying to get my head around JAVA.
How do I print out the content of the reference and not just their postition ? My program is ment to get some text in from the user, and print it out in a reverse order.
Here is my program (so far):
package myProgram;
import javax.swing.JOptionPane;
public class someRandomClass {
public static void main(String[] args) {
String word = JOptionPane.showInputDialog("Write som text here");
StringBuilder outPut = new StringBuilder();
for (int i = word.length()-1; i>=0; i--){
outPut.append(i);
}
System.out.println(outPut.toString());
}
}
I am greatfull for any help and tips! :)

In the line
outPut.append(i);
you are appending the value of your loop counter. You surely mean
outPut.append(word.charAt(i));

You seem to appending the integers instead of the appropriate characters. Try this instead:
outPut.append(word.substring(i, i + 1))
This way, the individual characters of word are appended to your StringBuilder. Note that the append method could also take a char as an argument, so you are also able to use word.charAt(i).

So, you want to emit the character at the position? Try using String.charAt.
outPut.append(word.charAt(i));
I'd probably avoid that and just index the char[] from String.toCharArray, though.
To be honest, I'd avoid doing the reversal loop manually to begin with... try something as follows:
final String word = JOptionPane.showInputDialog("Enter text below");
System.out.println(new StringBuilder(word).reverse());
StringBuilder.reverse should do the work for you (likely in a more efficient way, too). You also don't need to call toString manually, as println will do that for you.

String.split() Not Acting on Semicolon or Space Delimiters

This may be a simple question, but I have been Googling for over an hour and haven't found an answer yet.
I'm trying to simply use the String.split() method with a small Android application to split an input string. The input string will be something along the lines of: "Launch ip:192.168.1.101;port:5900". I'm doing this in two iterations to ensure that all of the required parameters are there. I'm first trying to do a split on spaces and semicolons to get the individual tokens sorted out. Next, I'm trying to split on colons in order to strip off the identification tags of each piece of information.
So, for example, I would expect the first round of split to give me the following data from the above example string:
(1) Launch
(2) ip:192.168.1.101
(3) port:5900
Then the second round would give me the following:
(1) 192.168.1.101
(2) 5900
However, the following code that I wrote doesn't give me what's expected:
private String[] splitString(String inputString)
{
String[] parsedString;
String[] orderedString = new String[SOSLauncherConstants.SOCKET_INPUT_STRING_PARSE_VALUE];
parsedString = inputString.trim().split("; ");
Log.i("info", "The parsed data is as follows for the initially parsed string of size " + parsedString.length + ": ");
for (int i = 0; i < parsedString.length; ++i)
{
Log.i("info", parsedString[i]);
}
for (int i = 0; i < parsedString.length; ++i )
{
if (parsedString[i].toLowerCase().contains(SOSLauncherConstants.PARSED_LAUNCH_COMMAND_VALUE))
{
orderedString[SOSLauncherConstants.PARSED_COMMAND_WORD] = parsedString[i];
}
if (parsedString[i].toLowerCase().contains("ip"))
{
orderedString[SOSLauncherConstants.PARSED_IP_VALUE] = parsedString[i].split(":")[1];
}
else if (parsedString[i].toLowerCase().contains("port"))
{
orderedString[SOSLauncherConstants.PARSED_PORT_VALUE] = parsedString[i].split(":")[1];
}
else if (parsedString[i].toLowerCase().contains("username"))
{
orderedString[SOSLauncherConstants.PARSED_USERNAME_VALUE] = parsedString[i].split(":")[1];
}
else if (parsedString[i].toLowerCase().contains("password"))
{
orderedString[SOSLauncherConstants.PARSED_PASSWORD_VALUE] = parsedString[i].split(":")[1];
}
else if (parsedString[i].toLowerCase().contains("color"))
{
orderedString[SOSLauncherConstants.PARSED_COLOR_VALUE] = parsedString[i].split(":")[1];
}
}
Log.i("info", "The parsed data is as follows for the second parsed string of size " + orderedString.length + ": ");
for (int i = 0; i < orderedString.length; ++i)
{
Log.i("info", orderedString[i]);
}
return orderedString;
}
For a result, I'm getting the following:
The parsed data is as follows for the parsed string of size 1:
launch ip:192.168.1.106;port:5900
The parsed data is as follows for the second parsed string of size 6:
launch ip:192.168.1.106;port:5900
192.168.1.106;port
And then, of course, it crashes because the for loop runs into a null string.
Side Note:
The following snippet is from the constants class that defines all of the string indexes --
public static final int SOCKET_INPUT_STRING_PARSE_VALUE = 6;
public static final int PARSED_COMMAND_WORD = 0;
public static final String PARSED_LAUNCH_COMMAND_VALUE = "launch";
public static final int PARSED_IP_VALUE = 1;
public static final int PARSED_PORT_VALUE = 2;
public static final int PARSED_USERNAME_VALUE = 3;
public static final int PARSED_PASSWORD_VALUE = 4;
public static final int PARSED_COLOR_VALUE = 5;
I looked into needing a possible escape (by inserting a \\ before the semicolon) on the semicolon delimiter, and even tried using it, but that didn't work. The odd part is that neither the space nor the semicolon function as a delimiter, yet the colon works on the second time around. Does anybody have any ideas what would cause this?
Thanks for your time!
EDIT: I should also add that I'm receiving the string over a WiFi socket connection. I don't think this should make a difference, but I'd like you to have all of the information that you need.

String.split(String) takes a regex. Use "[; ]". eg:
"foo;bar baz".split("[; ]")
will return an array containing "foo", "bar" and "baz".
If you need groups of spaces to work as a single delimiter, you can use something like:
"foo;bar baz".split("(;| +)")

I believe String.split() tries to split on each of the characters you specify together (or on a regex), not each character individually. That is, split(";.") would not split "a;b.c" at all, but would split "a;.b".
You may have better luck with Guava's Splitter, which is meant to be slightly less unpredictable than java.lang.String.split.
I would write something like
Iterable<String> splits = Splitter.on(CharMatcher.anyOf("; ")).split(string);
but Splitter also provides fluent-style customization like "trim results" or "skip over empty strings."

Is there a reason why you are using String.split(), but not using Regular Expressions? This is a perfect candidate for regex'es, esp if the string format is consistent.
I'm not sure if your format is fixed, and if it is, then the following regex should break it down for you (am sure that someone can come up with an even more elegant regex). If you have several command strings that follow, then you can use a more flexible regex and loop over all the groups:
Pattern p = Pattern.compile("([\w]*)[ ;](([\w]*):([^ ;]*))*");
Matcher m = p.match( <input string>);
if( m.find() )
command = m.group(1);
do{
id = m.group(3);
value = m.group(4);
} while( m.find() );
A great place to test out regex'es online is http://www.regexplanet.com/simple/index.html. It allows you to play with the regex without having to compile and launch you app every time if you just want to get the regex correct.