Long Form words to short form - java

How can I use indexOf() and CharAt() method to show the output like that?
"B R A M"
public class Test {
public static void main(String[] args) {
String s = new String("Business Requirement And Management");
String shortForm = "";
int index = 0;
int i=0;
while(i<s.length()) {
if(s.charAt(i) == ' '){
index = s.indexOf(" ");
shortForm += s.charAt(index+1);
}
i++;
}
System.out.println(shortForm);
}
}

I would use something like this:
StringBuilder sb = new StringBuilder();
String s = "Business Requirement And Management";
String[] splitted = s.split("\\s");
for(String str : splitted)
{
sb.append(str.charAt(0)).append(" ");
}
System.out.println(sb.toString());
Explanation:
We spilt the String at every whitespace character by calling split("\s") and store the new "substrings" in an array. Now iterate over the array and take the first character of each string and append it to the String.
Optionally you could use System.out.println(sb.toString().toUpperCase()) if you want to be certain that the initials are in upper case.

public class Test {
public static void main(String[] args) {
String s = new String("Business Requirement And Management");
String shortForm = "";
int index = 0;
while(s.length() > 0) {
index = s.indexOf(" ");
if(index == -1 || index == s.length() -1) break;
shortForm += s.charAt(index+1);
s = s.substring(index+1);
}
System.out.println(shortForm);
}
}

Related

parsing/converting task with characters and numbers within

It is necessary to repeat the character, as many times as the number behind it.
They are positive integer numbers.
case #1
input: "abc3leson11"
output: "abccclesonnnnnnnnnnn"
I already finish it in the following way:
String a = "abbc2kd3ijkl40ggg2H5uu";
String s = a + "*";
String numS = "";
int cnt = 0;
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
if (Character.isDigit(ch)) {
numS = numS + ch;
cnt++;
} else {
cnt++;
try {
for (int j = 0; j < Integer.parseInt(numS); j++) {
System.out.print(s.charAt(i - cnt));
}
if (i != s.length() - 1 && !Character.isDigit(s.charAt(i + 1))) {
System.out.print(s.charAt(i));
}
} catch (Exception e) {
if (i != s.length() - 1 && !Character.isDigit(s.charAt(i + 1))) {
System.out.print(s.charAt(i));
}
}
cnt = 0;
numS = "";
}
}
But I wonder is there some better solution with less and cleaner code?
Could you take a look below? I'm using a library from StringUtils from Apache Common Utils to repeat character:
public class MicsTest {
public static void main(String[] args) {
String input = "abc3leson11";
String output = input;
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(input);
while (m.find()) {
int number = Integer.valueOf(m.group());
char repeatedChar = input.charAt(m.start()-1);
output = output.replaceFirst(m.group(), StringUtils.repeat(repeatedChar, number));
}
System.out.println(output);
}
}
In case you don't want to use StringUtils. You can use the below custom method to achieve the same effect:
public static String repeat(char c, int times) {
char[] chars = new char[times];
Arrays.fill(chars, c);
return new String(chars);
}
Using java basic string regx should make it more terse as follows:
public class He1 {
private static final Pattern pattern = Pattern.compile("[a-zA-Z]+(\\d+).*");
// match the number between or the last using regx;
public static void main(String... args) {
String s = "abc3leson11";
System.out.println(parse(s));
s = "abbc2kd3ijkl40ggg2H5uu";
System.out.println(parse(s));
}
private static String parse(String s) {
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
int num = Integer.valueOf(matcher.group(1));
char prev = s.charAt(s.indexOf(String.valueOf(num)) - 1);
// locate the char before the number;
String repeated = new String(new char[num-1]).replace('\0', prev);
// since the prev is not deleted, we have to decrement the repeating number by 1;
s = s.replaceFirst(String.valueOf(num), repeated);
matcher = pattern.matcher(s);
}
return s;
}
}
And the output should be:
abccclesonnnnnnnnnnn
abbcckdddijkllllllllllllllllllllllllllllllllllllllllggggHHHHHuu
String g(String a){
String result = "";
String[] array = a.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)");
//System.out.println(java.util.Arrays.toString(array));
for(int i=0; i<array.length; i++){
String part = array[i];
result += part;
if(++i == array.length){
break;
}
char charToRepeat = part.charAt(part.length() - 1);
result += repeat(charToRepeat+"", new Integer(array[i]) - 1);
}
return result;
}
// In Java 11 this could be removed and replaced with the builtin `str.repeat(amount)`
String repeat(String str, int amount){
return new String(new char[amount]).replace("\0", str);
}
Try it online.
Explanation:
The split will split the letters and numbers:
abbc2kd3ijkl40ggg2H5uu would become ["abbc", "2", "kd", "3", "ijkl", "40", "ggg", "2", "H", "5", "uu"]
We then loop over the parts and add any strings as is to the result.
We then increase i by 1 first and if we're done (after the "uu") in the array above, it will break the loop.
If not the increase of i will put us at a number. So it will repeat the last character of the part x amount of times, where x is the number we found minus 1.
Here is another solution:
String str = "abbc2kd3ijkl40ggg2H5uu";
String[] part = str.split("(?<=\\d)(?=\\D)|(?=\\d)(?<=\\D)");
String res = "";
for(int i=0; i < part.length; i++){
if(i%2 == 0){
res = res + part[i];
}else {
res = res + StringUtils.repeat(part[i-1].charAt(part[i-1].length()-1),Integer.parseInt(part[i])-1);
}
}
System.out.println(res);
Yet another solution :
public static String getCustomizedString(String input) {
ArrayList<String > letters = new ArrayList<>(Arrays.asList(input.split("(\\d)")));
letters.removeAll(Arrays.asList(""));
ArrayList<String > digits = new ArrayList<>(Arrays.asList(input.split("(\\D)")));
digits.removeAll(Arrays.asList(""));
for(int i=0; i< digits.size(); i++) {
int iteration = Integer.valueOf(digits.get(i));
String letter = letters.get(i);
char c = letter.charAt(letter.length()-1);
for (int j = 0; j<iteration -1 ; j++) {
letters.set(i,letters.get(i).concat(String.valueOf(c)));
}
}
String finalResult = "";
for (String str : letters) {
finalResult += str;
}
return finalResult;
}
The usage:
public static void main(String[] args) {
String testString1 = "abbc2kd3ijkl40ggg2H5uu";
String testString2 = "abc3leson11";
System.out.println(getCustomizedString(testString1));
System.out.println(getCustomizedString(testString2));
}
And the result:
abbcckdddijkllllllllllllllllllllllllllllllllllllllllggggHHHHHuu
abccclesonnnnnnnnnnn

Reversing some words in a string

I need to reverse 5 or more character long words in a given string. For example:
* Given string: My name is Michael.
* Output: My name is leahciM.
Rest of the sentence stays the same, just those long words get reversed.
So far I came up with this:
public static String spinWords(String sentence) {
String[] splitWords = sentence.split(" ");
String reversedSentence = "";
String reversedWord = "";
for (String str : splitWords) {
if (str.length() >= 5) {
for (int i = str.length() - 1; i >= 0; i--)
reversedWord += (str.charAt(i) + " ");
}
}
}
And I have reversed those words, but
1) they are in one string, without a space
2) I dont know how to put them back into their places in string
Here is a suggestion:
write a method that reverses a string:
private static String reverse(String s) { ... }
then in your main method, call it when necessary:
if (str.length() >= 5) str = reverse(str);
you then need to put the words back together, presumably into the reversedSentence string:
reversedSentence += str + " "; //you will have an extra space at the end
Side notes:
using a StringBuilder may prove more efficient than string concatenation for longer sentences.
you could put all the words back into a List<String> within the loop and call reversedSentence = String.join(" ", list) after the loop
reversing a string can be done in one line - you should find numerous related Q&As on stackoverflow.
You can use StringBuilder
public static String spinWords(String sentence) {
String[] splitWords = sentence.split(" ");
StringBuilder builder = new StringBuilder();
for (String str : splitWords) {
if (str.length() < 5) {
builder.append(str);
else
builder.append(new StringBuilder(str).reverse().toString());
builder.append(" ");
}
return builder.toString().trim();
}
No need to use anything else you almost had it, just check your "for" loops and remember to add the unreversed string.
public static String spinWords(String sentence) {
String[] splitWords = sentence.split(" ");
String reversedSentence = "";
String reversedWord;
for (String str : splitWords) {
if (str.length() >= 5) {
reversedWord = "";
for (int i = str.length() - 1; i >= 0; i--) {
reversedWord += (str.charAt(i));
}
reversedSentence += " " + reversedWord;
} else {
reversedSentence += " " + str;
}
}
return reversedSentence;
}
Use StringBuilder to build the answer as you process the elements in splitWords.
You may also find the idiom of space with special first-time value (being "") useful.
There was also a bug in your original code.
So here is what I would do:
public class ReverseLongWord {
public static void main(String[] args) {
String testInput = "My name is Michael";
System.out.println(spinWords(testInput));
}
public static String spinWords(String sentence) {
String[] splitWords = sentence.split(" ");
String reversedSentence = "";
StringBuilder sb = new StringBuilder();
String space = ""; // first time special
String reversedWord = "";
for (String str : splitWords) {
if (str.length() >= 5) {
for (int i = str.length() - 1; i >= 0; i--) {
reversedWord += (str.charAt(i)); // Bug fixed
}
sb.append(space + reversedWord);
} else {
sb.append(space + str);
}
space = " "; // second time and onwards
}
return sb.toString();
}
}
The output of this program is the following, as you have specified:
My name is leahciM
I think the reverse method as some people suggest would be the easiest way, here I share my implementation
public static void main(String[] args) {
System.out.println(concatenatePhrase("My name is Michael"));
System.out.println(concatenatePhrase("Some randoms words with differents sizes and random words"));
}
private static String concatenatePhrase(String phrase) {
StringBuilder completePhrase = new StringBuilder();
String[] phrases = phrase.split(" ");
for (String word : phrases) {
if (word.length() >= 5) {
completePhrase.append(reverseWord(word).append(" "));
} else {
completePhrase.append(word).append(" ");
}
}
return completePhrase.toString().trim();
}
private static StringBuilder reverseWord(String wordPassed) {
StringBuilder word = new StringBuilder(wordPassed);
return word.reverse();
}

Replace two strings without overriding the other value

I have an assignment for school due at midnight today. I have finished almost all the assignment except for one question. I need to swap "r" and "q" with each other as values. So, if you enter "r" in the compiler you should get "q" if you enter "q" you get "r"(Using JOptionPane). For example, if your name is Quart, the compiler should print Ruaqt. I tried using the replace.All method, but once I can only swap "r" or "q" not both. I know I need a temporary variable, but do not know anything else...
We had to replace vowels with the letter after them so I did this:
String firstName = JOptionPane
.showInputDialog("What is your first name?");
String lastName = JOptionPane
.showInputDialog("What is your last name?");
String fullname = firstname + lastname;
String lowername = fullName.toLowerCase();
String encryptedname = lowername.replaceAll("a", "b")
.replaceAll("e", "f").replaceAll("i", "j").replaceAll("o", "p")
.replaceAll("u", "v");
Thanks
Dunno why the 2 answers using StringBuilder are both making the thing more complicated than needed.
Here is the way you can use StringBuilder to do that single character swap:
public static String swapChar(String string, char c1, char c2) {
StringBuilder sb = new StringBuilder(string);
for (int i = 0; i < sb.length(); ++i) {
if (sb.charAt(i) == c1) {
sb.setCharAt(i, c2);
} else if (sb.charAt(i) == c2) {
sb.setCharAt(i, c1);
}
}
return sb.toString();
}
Update :
Just found that what you are looking for is actually doing a bunch of replace of character at the same time. That can be cleanly done by providing a Map as parameter:
public static String replaceChars(String string, Map<Character,Character> cmap) {
StringBuilder sb = new StringBuilder(string);
for (int i = 0; i < sb.length(); ++i) {
if (cmap.containsKey(sb.charAt(i)) {
sb.setCharAt(i, cmap.get(sb.charAt(i));
}
}
return sb.toString();
}
to use it:
// or make a util method to make these even easier to create
Map<Character,Character> cmap = new HashMap<Character,Character>();
cmap.put('r','q');
cmap.put('q','r');
cmap.put('a','b');
cmap.put('e','f');
cmap.put('i','j');
cmap.put('o','p');
cmap.put('u','v');
and simply do a replace:
String result = replaceChars(inputString, cmap);
or even simpler, by making use of Apache Commons Lang:
String result = StringUtils.replaceChars(inputString, "rqaeiou", "qrbfjpv");
You can try this.
private static final char Q_STR = 'q';
private static final char R_STR = 'r';
public static String replaceString(String original, int position, char strToReplace) {
StringBuilder strBuilder = new StringBuilder(original);
if (strToReplace == Q_STR) {
strBuilder.setCharAt(position, R_STR);
} else if (strToReplace == R_STR){
strBuilder.setCharAt(position, Q_STR);
}
return strBuilder.toString();
}
public static void main(String[] args) {
String firstname = "Quart";
String lastname = " QuartLastName";
String fullname = firstname + lastname;
String lowername = fullname.toLowerCase();
//get all chars in String
char[] array = lowername.toCharArray();
//list to keep original position of Q char
List<Integer> allQPosition = new ArrayList<Integer>();
//list to keep original position of R char
List<Integer> allRPosition = new ArrayList<Integer>();
for (int i = 0; i < array.length; i++) {
if(array[i] == 'q') {
allQPosition.add(i);
} else if(array[i] == 'r') {
allRPosition.add(i);
}
}
//replace q
for (Integer integer : allQPosition) {
lowername = replaceString(lowername, integer, Q_STR);
}
//replace r
for (Integer integer : allRPosition) {
lowername = replaceString(lowername, integer, R_STR);
}
//replace others
String encryptedname = lowername.replace("a", "b")
.replace("e", "f")
.replace("i", "j")
.replace("o", "p")
.replace("u", "v");
System.out.println("Result: " + encryptedname);
}
My solution is:
Keep all position of 'q' and 'r' from original String.
Replace each of them
Replace the rest of other chars
Hope this help
public static void main(String[] args) {
String origin = "r and q";
System.out.println(newReplacement(origin, 'r', 'q'));
}
private static String newReplacement(String origin, char firstChar, char secondChar) {
StringBuffer stringBuffer = new StringBuffer(origin);
for(int i = 0; i < origin.length(); i++) {
if(origin.charAt(i) == firstChar) {
stringBuffer.replace(i, i+1, secondChar + "");
continue;
}
if(origin.charAt(i) == secondChar) {
stringBuffer.replace(i, i+1, firstChar + "");
}
}
return stringBuffer.toString();
}
Rewrite replace method with simple one.

I want to split string without using split function?

I want to split string without using split . can anybody solve my problem I am tried but
I cannot find the exact logic.
Since this seems to be a task designed as coding practice, I'll only guide. No code for you, sir, though the logic and the code aren't that far separated.
You will need to loop through each character of the string, and determine whether or not the character is the delimiter (comma or semicolon, for instance). If not, add it to the last element of the array you plan to return. If it is the delimiter, create a new empty string as the array's last element to start feeding your characters into.
I'm going to assume that this is homework, so I will only give snippets as hints:
Finding indices of all occurrences of a given substring
Here's an example of using indexOf with the fromIndex parameter to find all occurrences of a substring within a larger string:
String text = "012ab567ab0123ab";
// finding all occurrences forward: Method #1
for (int i = text.indexOf("ab"); i != -1; i = text.indexOf("ab", i+1)) {
System.out.println(i);
} // prints "3", "8", "14"
// finding all occurrences forward: Method #2
for (int i = -1; (i = text.indexOf("ab", i+1)) != -1; ) {
System.out.println(i);
} // prints "3", "8", "14"
String API links
int indexOf(String, int fromIndex)
Returns the index within this string of the first occurrence of the specified substring, starting at the specified index. If no such occurrence exists, -1 is returned.
Related questions
Searching for one string in another string
Extracting substrings at given indices out of a string
This snippet extracts substring at given indices out of a string and puts them into a List<String>:
String text = "0123456789abcdefghij";
List<String> parts = new ArrayList<String>();
parts.add(text.substring(0, 5));
parts.add(text.substring(3, 7));
parts.add(text.substring(9, 13));
parts.add(text.substring(18, 20));
System.out.println(parts); // prints "[01234, 3456, 9abc, ij]"
String[] partsArray = parts.toArray(new String[0]);
Some key ideas:
Effective Java 2nd Edition, Item 25: Prefer lists to arrays
Works especially nicely if you don't know how many parts there'll be in advance
String API links
String substring(int beginIndex, int endIndex)
Returns a new string that is a substring of this string. The substring begins at the specified beginIndex and extends to the character at index endIndex - 1.
Related questions
Fill array with List data
You do now that most of the java standard libraries are open source
In this case you can start here
Use String tokenizer to split strings in Java without split:
import java.util.StringTokenizer;
public class tt {
public static void main(String a[]){
String s = "012ab567ab0123ab";
String delims = "ab ";
StringTokenizer st = new StringTokenizer(s, delims);
System.out.println("No of Token = " + st.countTokens());
while (st.hasMoreTokens()) {
System.out.println(st.nextToken());
}
}
}
This is the right answer
import java.util.StringTokenizer;
public class tt {
public static void main(String a[]){
String s = "012ab567ab0123ab";
String delims = "ab ";
StringTokenizer st = new StringTokenizer(s, delims);
System.out.println("No of Token = " + st.countTokens());
while (st.hasMoreTokens())
{
System.out.println(st.nextToken());
}
}
}
/**
* My method split without javas split.
* Return array with words after mySplit from two texts;
* Uses trim.
*/
public class NoJavaSplit {
public static void main(String[] args) {
String text1 = "Some text for example ";
String text2 = " Second sentences ";
System.out.println(Arrays.toString(mySplit(text1, text2)));
}
private static String [] mySplit(String text1, String text2) {
text1 = text1.trim() + " " + text2.trim() + " ";
char n = ' ';
int massValue = 0;
for (int i = 0; i < text1.length(); i++) {
if (text1.charAt(i) == n) {
massValue++;
}
}
String[] splitArray = new String[massValue];
for (int i = 0; i < splitArray.length; ) {
for (int j = 0; j < text1.length(); j++) {
if (text1.charAt(j) == n) {
splitArray[i] = text1.substring(0, j);
text1 = text1.substring(j + 1, text1.length());
j = 0;
i++;
}
}
return splitArray;
}
return null;
}
}
you can try, the way i did `{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
for(int i = 0; i <str.length();i++) {
if(str.charAt(i)==' ') { // whenever it found space it'll create separate words from string
System.out.println();
continue;
}
System.out.print(str.charAt(i));
}
sc.close();
}`
The logic is: go through the whole string starting from first character and whenever you find a space copy the last part to a new string.. not that hard?
The way to go is to define the function you need first. In this case, it would probably be:
String[] split(String s, String separator)
The return type doesn't have to be an array. It can also be a list:
List<String> split(String s, String separator)
The code would then be roughly as follows:
start at the beginning
find the next occurence of the delimiter
the substring between the end of the previous delimiter and the start of the current delimiter is added to the result
continue with step 2 until you have reached the end of the string
There are many fine points that you need to consider:
What happens if the string starts or ends with the delimiter?
What if multiple delimiters appear next to each other?
What should be the result of splitting the empty string? (1 empty field or 0 fields)
You can do it using Java standard libraries.
Say the delimiter is : and
String s = "Harry:Potter"
int a = s.find(delimiter);
and then add
s.substring(start, a)
to a new String array.
Keep doing this till your start < string length
Should be enough I guess.
public class MySplit {
public static String[] mySplit(String text,String delemeter){
java.util.List<String> parts = new java.util.ArrayList<String>();
text+=delemeter;
for (int i = text.indexOf(delemeter), j=0; i != -1;) {
parts.add(text.substring(j,i));
j=i+delemeter.length();
i = text.indexOf(delemeter,j);
}
return parts.toArray(new String[0]);
}
public static void main(String[] args) {
String str="012ab567ab0123ab";
String delemeter="ab";
String result[]=mySplit(str,delemeter);
for(String s:result)
System.out.println(s);
}
}
public class WithoutSpit_method {
public static void main(String arg[])
{
char[]str;
String s="Computer_software_developer_gautam";
String s1[];
for(int i=0;i<s.length()-1;)
{
int lengh=s.indexOf("_",i);
if(lengh==-1)
{
lengh=s.length();
}
System.out.print(" "+s.substring(i,lengh));
i=lengh+1;
}
}
}
Result: Computer software developer gautam
Here is my way of doing with Scanner;
import java.util.Scanner;
public class spilt {
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter the String to be Spilted : ");
String st = input.nextLine();
Scanner str = new Scanner(st);
while (str.hasNext())
{
System.out.println(str.next());
}
}
}
Hope it Helps!!!!!
public class StringWitoutPre {
public static void main(String[] args) {
String str = "md taufique reja";
int len = str.length();
char ch[] = str.toCharArray();
String tmp = " ";
boolean flag = false;
for (int i = 0; i < str.length(); i++) {
if (ch[i] != ' ') {
tmp = tmp + ch[i];
flag = false;
} else {
flag = true;
}
if (flag || i == len - 1) {
System.out.println(tmp);
tmp = " ";
}
}
}
}
In Java8 we can use Pattern and get the things done in more easy way. Here is the code.
package com.company;
import java.util.regex.Pattern;
public class umeshtest {
public static void main(String a[]) {
String ss = "I'm Testing and testing the new feature";
Pattern.compile(" ").splitAsStream(ss).forEach(s -> System.out.println(s));
}
}
static void splitString(String s, int index) {
char[] firstPart = new char[index];
char[] secondPart = new char[s.length() - index];
int j = 0;
for (int i = 0; i < s.length(); i++) {
if (i < index) {
firstPart[i] = s.charAt(i);
} else {
secondPart[j] = s.charAt(i);
if (j < s.length()-index) {
j++;
}
}
}
System.out.println(firstPart);
System.out.println(secondPart);
}
import java.util.Scanner;
public class Split {
static Scanner in = new Scanner(System.in);
static void printArray(String[] array){
for (int i = 0; i < array.length; i++) {
if(i!=array.length-1)
System.out.print(array[i]+",");
else
System.out.println(array[i]);
}
}
static String delimeterTrim(String str){
char ch = str.charAt(str.length()-1);
if(ch=='.'||ch=='!'||ch==';'){
str = str.substring(0,str.length()-1);
}
return str;
}
private static String [] mySplit(String text, char reg, boolean delimiterTrim) {
if(delimiterTrim){
text = delimeterTrim(text);
}
text = text.trim() + " ";
int massValue = 0;
for (int i = 0; i < text.length(); i++) {
if (text.charAt(i) == reg) {
massValue++;
}
}
String[] splitArray = new String[massValue];
for (int i = 0; i < splitArray.length; ) {
for (int j = 0; j < text.length(); j++) {
if (text.charAt(j) == reg) {
splitArray[i] = text.substring(0, j);
text = text.substring(j + 1, text.length());
j = 0;
i++;
}
}
return splitArray;
}
return null;
}
public static void main(String[] args) {
System.out.println("Enter the sentence :");
String text = in.nextLine();
//System.out.println("Enter the regex character :");
//char regex = in.next().charAt(0);
System.out.println("Do you want to trim the delimeter ?");
String delch = in.next();
boolean ch = false;
if(delch.equalsIgnoreCase("yes")){
ch = true;
}
System.out.println("Output String array is : ");
printArray(mySplit(text,' ',ch));
}
}
Split a string without using split()
static String[] splitAString(String abc, char splitWith){
char[] ch=abc.toCharArray();
String temp="";
int j=0,length=0,size=0;
for(int i=0;i<abc.length();i++){
if(splitWith==abc.charAt(i)){
size++;
}
}
String[] arr=new String[size+1];
for(int i=0;i<ch.length;i++){
if(length>j){
j++;
temp="";
}
if(splitWith==ch[i]){
length++;
}else{
temp +=Character.toString(ch[i]);
}
arr[j]=temp;
}
return arr;
}
public static void main(String[] args) {
String[] arr=splitAString("abc-efg-ijk", '-');
for(int i=0;i<arr.length;i++){
System.out.println(arr[i]);
}
}
}
You cant split with out using split(). Your only other option is to get the strings char indexes and and get sub strings.

How to upper case every first letter of word in a string? [duplicate]

This question already has answers here:
How to capitalize the first character of each word in a string
(51 answers)
Closed 3 years ago.
I have a string: "hello good old world" and i want to upper case every first letter of every word, not the whole string with .toUpperCase(). Is there an existing java helper which does the job?
Have a look at ACL WordUtils.
WordUtils.capitalize("your string") == "Your String"
Here is the code
String source = "hello good old world";
StringBuffer res = new StringBuffer();
String[] strArr = source.split(" ");
for (String str : strArr) {
char[] stringArray = str.trim().toCharArray();
stringArray[0] = Character.toUpperCase(stringArray[0]);
str = new String(stringArray);
res.append(str).append(" ");
}
System.out.print("Result: " + res.toString().trim());
sString = sString.toLowerCase();
sString = Character.toString(sString.charAt(0)).toUpperCase()+sString.substring(1);
i dont know if there is a function but this would do the job in case there is no exsiting one:
String s = "here are a bunch of words";
final StringBuilder result = new StringBuilder(s.length());
String[] words = s.split("\\s");
for(int i=0,l=words.length;i<l;++i) {
if(i>0) result.append(" ");
result.append(Character.toUpperCase(words[i].charAt(0)))
.append(words[i].substring(1));
}
import org.apache.commons.lang.WordUtils;
public class CapitalizeFirstLetterInString {
public static void main(String[] args) {
// only the first letter of each word is capitalized.
String wordStr = WordUtils.capitalize("this is first WORD capital test.");
//Capitalize method capitalizes only first character of a String
System.out.println("wordStr= " + wordStr);
wordStr = WordUtils.capitalizeFully("this is first WORD capital test.");
// This method capitalizes first character of a String and make rest of the characters lowercase
System.out.println("wordStr = " + wordStr );
}
}
Output :
This Is First WORD Capital Test.
This Is First Word Capital Test.
Here's a very simple, compact solution. str contains the variable of whatever you want to do the upper case on.
StringBuilder b = new StringBuilder(str);
int i = 0;
do {
b.replace(i, i + 1, b.substring(i,i + 1).toUpperCase());
i = b.indexOf(" ", i) + 1;
} while (i > 0 && i < b.length());
System.out.println(b.toString());
It's best to work with StringBuilder because String is immutable and it's inefficient to generate new strings for each word.
Trying to be more memory efficient than splitting the string into multiple strings, and using the strategy shown by Darshana Sri Lanka. Also, handles all white space between words, not just the " " character.
public static String UppercaseFirstLetters(String str)
{
boolean prevWasWhiteSp = true;
char[] chars = str.toCharArray();
for (int i = 0; i < chars.length; i++) {
if (Character.isLetter(chars[i])) {
if (prevWasWhiteSp) {
chars[i] = Character.toUpperCase(chars[i]);
}
prevWasWhiteSp = false;
} else {
prevWasWhiteSp = Character.isWhitespace(chars[i]);
}
}
return new String(chars);
}
String s = "java is an object oriented programming language.";
final StringBuilder result = new StringBuilder(s.length());
String words[] = s.split("\\ "); // space found then split it
for (int i = 0; i < words.length; i++)
{
if (i > 0){
result.append(" ");
}
result.append(Character.toUpperCase(words[i].charAt(0))).append(
words[i].substring(1));
}
System.out.println(result);
Output: Java Is An Object Oriented Programming Language.
Also you can take a look into StringUtils library. It has a bunch of cool stuff.
My code after reading a few above answers.
/**
* Returns the given underscored_word_group as a Human Readable Word Group.
* (Underscores are replaced by spaces and capitalized following words.)
*
* #param pWord
* String to be made more readable
* #return Human-readable string
*/
public static String humanize2(String pWord)
{
StringBuilder sb = new StringBuilder();
String[] words = pWord.replaceAll("_", " ").split("\\s");
for (int i = 0; i < words.length; i++)
{
if (i > 0)
sb.append(" ");
if (words[i].length() > 0)
{
sb.append(Character.toUpperCase(words[i].charAt(0)));
if (words[i].length() > 1)
{
sb.append(words[i].substring(1));
}
}
}
return sb.toString();
}
import java.util.Scanner;
public class CapitolizeOneString {
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.print(" Please enter Your word = ");
String str=scan.nextLine();
printCapitalized( str );
} // end main()
static void printCapitalized( String str ) {
// Print a copy of str to standard output, with the
// first letter of each word in upper case.
char ch; // One of the characters in str.
char prevCh; // The character that comes before ch in the string.
int i; // A position in str, from 0 to str.length()-1.
prevCh = '.'; // Prime the loop with any non-letter character.
for ( i = 0; i < str.length(); i++ ) {
ch = str.charAt(i);
if ( Character.isLetter(ch) && ! Character.isLetter(prevCh) )
System.out.print( Character.toUpperCase(ch) );
else
System.out.print( ch );
prevCh = ch; // prevCh for next iteration is ch.
}
System.out.println();
}
} // end class
public class WordChangeInCapital{
public static void main(String[] args)
{
String s="this is string example";
System.out.println(s);
//this is input data.
//this example for a string where each word must be started in capital letter
StringBuffer sb=new StringBuffer(s);
int i=0;
do{
b.replace(i,i+1,sb.substring(i,i+1).toUpperCase());
i=b.indexOf(" ",i)+1;
} while(i>0 && i<sb.length());
System.out.println(sb.length());
}
}
package com.raj.samplestring;
/**
* #author gnagara
*/
public class SampleString {
/**
* #param args
*/
public static void main(String[] args) {
String[] stringArray;
String givenString = "ramu is Arr Good boy";
stringArray = givenString.split(" ");
for(int i=0; i<stringArray.length;i++){
if(!Character.isUpperCase(stringArray[i].charAt(0))){
Character c = stringArray[i].charAt(0);
Character change = Character.toUpperCase(c);
StringBuffer ss = new StringBuffer(stringArray[i]);
ss.insert(0, change);
ss.deleteCharAt(1);
stringArray[i]= ss.toString();
}
}
for(String e:stringArray){
System.out.println(e);
}
}
}
Here is an easy solution:
public class CapitalFirstLetters {
public static void main(String[] args) {
String word = "it's java, baby!";
String[] wordSplit;
String wordCapital = "";
wordSplit = word.split(" ");
for (int i = 0; i < wordSplit.length; i++) {
wordCapital = wordSplit[i].substring(0, 1).toUpperCase() + wordSplit[i].substring(1) + " ";
}
System.out.println(wordCapital);
}}
public String UpperCaseWords(String line)
{
line = line.trim().toLowerCase();
String data[] = line.split("\\s");
line = "";
for(int i =0;i< data.length;i++)
{
if(data[i].length()>1)
line = line + data[i].substring(0,1).toUpperCase()+data[i].substring(1)+" ";
else
line = line + data[i].toUpperCase();
}
return line.trim();
}
So much simpler with regexes:
Pattern spaces=Pattern.compile("\\s+[a-z]");
Matcher m=spaces.matcher(word);
StringBuilder capitalWordBuilder=new StringBuilder(word.substring(0,1).toUpperCase());
int prevStart=1;
while(m.find()) {
capitalWordBuilder.append(word.substring(prevStart,m.end()-1));
capitalWordBuilder.append(word.substring(m.end()-1,m.end()).toUpperCase());
prevStart=m.end();
}
capitalWordBuilder.append(word.substring(prevStart,word.length()));
Output for input: "this sentence Has Weird caps"
This Sentence Has Weird Caps

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