Develop Reference

String match of only 3 specific words

I want a regex to match a string that only has the words A,I and D without any order or sort
also:if the string has a letter thats not any of these then doesnt go into the if
I have tried with || and other symbols but still cant get it
Doesnt have to be a regex Im just trying to find a way to solve it
String message = "AIDDDAAIDAAA"
if(message.matches("(A|D|I)")){
System.out.println("Matches");
}

You can use include all characters you are interested in within a square bracket. To match one or more occurrences of these characters in square brackets, append + to it. The message string should be entirely made up of only these characters for it to be considered a match.
Try this.
String message = "ADAIAIAIAIAIADDDAI";
if(message .matches("[ADI]+")) {
System.out.println("Matches");
}

Since you're asking about "words", I guess A, D and I stand for words with more than one letter and that that is the reaason why you're not using the character class [ADI]. You just have to add a + because the message consists of more words.
String message = "AIDDDAAIDAAA";
if (message.matches("(A|D|I)+")) {
System.out.println("Matches");
}
scigs answer works as well.

You could use string replace to do something similar:
String message = "AIDDDAAIDAAA";
message = message.replace("A","")
.replace("I","")
.replace("D","");
if (message.equals("")) {
//do your thing
}

You could just check each char composing the String:
String message = "AIDDDAAIDAAA";
boolean matches = true;
for (int i=0; i<message.length(); i++;){
if (message.charAt(i)!='A' && message.charAt(i)!='I' && message.charAt(i)!='D'){
matches = false;
break;
}
}
if (matches) System.out.println("Matches");

Wrong Answer with Java

I'm a student that is learning Java, and I have this code:
lletres = lletres.replace(lletres.charAt(2), codi.charAt(codi.indexOf(lletres.charAt(2)) + 1));
lletres is a string, and it's like this
lletres = "BBB"
The result is "CCC" and I only want to change the last B, so the result can be like this: "BBC".

Reading the documentation for String.replace should explain what happened here (I marked the relevant part in bold):
Returns a string resulting from replacing all occurrences of oldChar in this string with newChar.
One way to solve it is to break the string up to the parts you want and then put it back together again. E.g.:
lletres = lletres.substring(0, 2) + (char)(lletres.charAt(2) + 1);

As others pointed replace() will replace all the occurrences which matched.
So, instead you can make use of replaceFirst() which will accept the regx
lletres = lletres.replaceFirst( lletres.charAt( 2 ) + "$", (char) ( lletres.charAt( 2 ) + 1 ) + "" )

You could use StringBuilder for your purpose:
String lletres = "BBB";
String codi = "CCC";
StringBuilder sb = new StringBuilder(lletres);
sb.setCharAt(2, codi.charAt(codi.indexOf(lletres.charAt(2)) + 1));
lletres = sb.toString();

If you need to change only the last occurrence in the string, you need to split the string into parts first. I hope following snippet will be helpful to you.
String lletres = "BBB";
int lastIndex = lletres.lastIndexOf('B');
lletres = lletres.substring(0, lastIndex) + 'C' + lletres.substring(lastIndex+1);
This code will find index of last letter B and stores it in lastIndex. Then it splits the string and replaces that B letter with C letter.
Please keep in mind that this snippet doesn't check whether or not the letter B is present in the string.
With slight modification you can get it to replace whole parts of the string, not only letters. :)

Try this one.
class Rplce
{
public static void main(String[] args)
{
String codi = "CCC";
String lletres = "BBB";
int char_no_to_be_replaced = 2;
lletres = lletres.substring(0,char_no_to_be_replaced ) + codi.charAt(codi.indexOf(lletres.charAt(char_no_to_be_replaced )) + 1) + lletres.substring(char_no_to_be_replaced + 1);
System.out.println(lletres);
}
}

use this to replace the last character
lletres = lletres.replaceAll(".{1}$", String.valueOf((char) (lletres.charAt(2) + 1)));

suppose you have dynamic value at last index and you want to replace that value will increasing one then use this code
String lletres = "BBB";
int atIndex = lletres.lastIndexOf('B');
char ReplacementChar = (char)(lletres.charAt(lletres.lastIndexOf('B'))+1);
lletres= lletres.substring(0, atIndex)+ReplacementChar;
System.out.println(lletres);
output
BBC

Remove "yak" from a string

I am working on a codingbat problem:Given a string, return a version where all the "yak" are removed, but the "a" can be any char. The "yak" strings will not overlap. I am look at the solution but there is one part of that code I do not understand.....
How come the first part of the if statement "i = i+ 2" could return the string and you don't need anything else? I mean after all these three conditions are met and just write i = i + 2,and that's it. that is going to give you a String as a result. I don't get it, please help.
public String stringYak(String str) {
String result = "";
for (int i=0; i<str.length(); i++) {
// Look for i starting a "yak" -- advance i in that case
if (i+2<str.length() && str.charAt(i)=='y' && str.charAt(i+2)=='k') {
i = i + 2;
} else { // Otherwise do the normal append
result = result + str.charAt(i);
}
}
return result;
}

The code is selecting what characters to put in the new string.
We go through the characters one by one.
If we run into a "y.k", skip this whole section
Else add the character to the new string.
[a][b][y][c][k][d] => New String: [a] (a is okay)
.|..........................
[a][b][y][c][k][d] => New String: [a][b] (b is okay)
......|.....................
[a][b][y][c][k][d] => New String: [a][b] (Oops! We have run into the y.k
pattern, skip it) ..........|.................
[a][b][y][c][k][d] => New String: [a][b][d] (d is okay)
.....................|......
Final String: [a][b][d]

This may not be the style of answer you want, but a simple call to String#replaceAll() should work:
String str = "Some string yak containing yok.";
str = str.replaceAll("y.k", "");

Doing i = i + 2 is not going to give you any String it is simply incrementing the for loop so you do not need to re-evaluate the next two chars as they have already been evaluated.
The key point is in the else part, which will append the char if it is not y and the character after-next is also not k

Get certain data from text - Java

I am creating a bukkit plugin for minecraft and i need to know a few things before i move on.
I want to check if a text has this layout: "B:10 S:5" for example.
It stands for Buy:amount and Sell:amount
How can i check the easiest way if it follows the syntax?
It can be any number that is 0 or over.
Another problem is to get this data out of the text. how can i check what text is after B: and S: and return it as an integer
I have not tried out this yet because i have no idea where to start.
Thanks for help!

In the simple problem you gave, you can get away with a simple answer. Otherwise, see the regex answer below.
boolean test(String str){
try{
//String str = "B:10 S:5";
String[] arr = str.split(" ");//split to left and right of space = [B:10,S:5]
String[] bArr = arr[0].split(":");//split ...first colon = [B,10]
String[] sArr = arr[1].split(":");//... second colon = [S,5]
//need to use try/catch here in case the string is not an int value.
String labelB = bArr[0];
Integer b = Integer.parseInt(bArr[1]);
String labelS = sArr[0];
Integer s = Integer.parseInt(sArr[1]);
}catch( Exception e){return false;}
return true;
}
See my answer here for a related task. More related details below.
How can I parse a string for a set?
Essentially, you need to use regex and iterate through the groups. Just in case the grammar is not always B and S, I made this more abstract.Also, if there are extra white spaces in the middle for what ever reason, I made that more broad too. The pattern says there are 4 groups (indicated by parentheses): label1, number1, label2, and number2. + means 1 or more. [] means a set of characters. a-z is a range of characters (don't put anything between A-Z and a-z). There are also other ways of showing alpha and numeric patterns, but these are easier to read.
//this is expensive
Pattern p=Pattern.compile("([A-Za-z]+):([0-9]+)[ ]+([A-Za-z]+):([0-9]+)");
boolean test(String txt){
Matcher m=p.matcher(txt);
if(!m.matches())return false;
int groups=m.groupCount();//should only equal 5 (default whole match+4 groups) here, but you can test this
System.out.println("Matched: " + m.group(0));
//Label1 = m.group(1);
//val1 = m.group(2);
//Label2 = m.group(3);
//val2 = m.group(4);
return true;
}

Use Regular Expression.
In your case,^B:(\d)+ S:(\d)+$ is enough.
In java, to use a regular expression:
public class RegExExample {
public static void main(String[] args) {
Pattern p = Pattern.compile("^B:(\d)+ S:(\d)+$");
for (int i = 0; i < args.length; i++)
if (p.matcher(args[i]).matches())
System.out.println( "ARGUMENT #" + i + " IS VALID!")
else
System.out.println( "ARGUMENT #" + i + " IS INVALID!");
}
}
This sample program take inputs from command line, validate it against the pattern and print the result to STDOUT.

Java: Next character in String

I have one String generated of random characters that will encrypt another String given by the user by adding the first character from the String with the first character of the given String. It's working fine, but if the user were to enter multiple words with spaces in between, I want to choose the next character of the first String rather than code the space itself. Is that possible? This is what I have:
(random is the coded string and sentenceUpper is string given by user)
public static void encrypt(String sentenceUpper){
String newSentence = "";
for(int i = 0; i < sentenceUpper.length(); i++){
char one = random.charAt(i);
char two = sentenceUpper.charAt(i);
if(one < 'A' || one > 'Z'){
two = sentenceUpper.charAt(1 + i);}
char result = (char)((one + two)%26 + 'A');
newSentence += "" + result;
}
EDIT FOR BETTER EXPLANATION:
I have:
String random = "WFAZYZAZOHS";
I would like to code user input:
String upperCase: "YOU GO";
So, I'm going to take Y + L = U, etc...
to get :
"UTUSEN
"
But I see that there's a space in "YOU GO" , So I'd like to change it to:
WFA ZY + YOU GO = UTU SE.
I hope that's better explained.

The simplest way to do this would probably be to use an if statement to run the code in the loop only if the character is not a space. If you don't want to skip the character in the random string, you would need a separate variable to track the current character index in that string.
Example: Put this after defining one and two and put the rest of the loop inside it:
if(two==' '){
...
}
Then, add the space in the output:
else{
newSentence+=" ";
}