As you know, dropWhile for streams was introduced in java 9. But if the target of the project is java 8, you can't use it.
Example code:
public static String getParameterValueOrDefault(String[] args, String paramName, String defaultVal) {
return Arrays.stream(args).sequential().dropWhile(arg->!arg.equals("/"+paramName) && !arg.equals("-"+paramName)).skip(1).findFirst().orElseGet(()->defaultVal);
}
What I want is an equivalent lambda expression written in java 8.
You can use
public static String getParameterValueOrDefault(
String[] args, String paramName, String defaultVal) {
int ix = IntStream.range(0, args.length)
.filter(i -> args[i].matches("[/-]" + Pattern.quote(paramName)))
.map(i -> i + 1)
.findFirst().orElse(args.length);
return ix < args.length? args[ix]: defaultVal;
}
The matches approach is for convenience, if you prefer compact code. If you want an efficient check, you may instead use:
public static String getParameterValueOrDefault(
String[] args, String paramName, String defaultVal) {
int ix = IntStream.range(0, args.length)
.filter(i -> {
String arg = args[i];
return arg.length() == paramName.length() + 1 && arg.endsWith(paramName)
&& (arg.charAt(0) == '-' || arg.charAt(0) == '/');
})
.map(i -> i + 1)
.findFirst().orElse(args.length);
return ix < args.length? args[ix]: defaultVal;
}
I found this way:
public static String getParameterValueOrDefault(String[] args, String paramName, String defaultVal) {
MutableBoolean foundParam = new MutableBoolean(false);
return Arrays.stream(args).sequential().peek(arg->foundParam.value = foundParam.value || arg.equals("/"+paramName) || arg.equals("-"+paramName)).filter(arg->foundParam.value).skip(1).findFirst().orElseGet(()->defaultVal);
}
Where MutableBoolean is:
private static class MutableBoolean {
boolean value;
public MutableBoolean(boolean value) {
this.value = value;
}
}
I've read the documentation and I run into this issue of using the spread operator with an ArrayList Collection, and I want to know how to solve the mismatch type or implement a way to use it with ArrayList
I'll attach an image of the code along with the code.
fun howSum(targetSum: Int, numbers: ArrayList<Int>): ArrayList<Int>? {
if (targetSum == 0) return arrayListOf();
if (targetSum < 0) return null;
for (number: Int in numbers){
val remainder = targetSum - number;
val remainderResult = howSum(remainder, numbers);
if (remainderResult != null){
return arrayListOf(*remainderResult, number)
}
}
return null
}
Any comment could be helpful...
I think you need to give us more information about what you are trying to do for a better answer.
The spread operator is for passing an array in place of a varargs argument, but you can't add additional arguments to the array at the same time.
If you want a new ArrayList that contains the contents of another ArrayList with an extra element added, you can do something like this:
fun main() {
val foo = arrayListOf(1, 2, 3)
val bar = arrayListOf<Int>().apply {
addAll(foo)
add(4)
}
println(foo)
println(bar)
}
Output:
[1, 2, 3]
[1, 2, 3, 4]
But it's not efficient, because it copies all the items of foo into bar.
Spread operator is not applicable to Lists, it's intended only for arrays:
fun howSum(targetSum: Int, numbers: ArrayList<Int>): IntArray? {
if (targetSum == 0) return intArrayOf()
if (targetSum < 0) return null
for (number: Int in numbers) {
val remainder = targetSum - number;
val remainderResult = howSum(remainder, numbers);
if (remainderResult != null) {
return intArrayOf(*remainderResult, number)
}
}
return null
}
If you want to create new List of the other one with addition of some element, you can use + operator:
fun howSum(targetSum: Int, numbers: ArrayList<Int>): List<Int>? {
if (targetSum == 0) return arrayListOf();
if (targetSum < 0) return null;
for (number: Int in numbers){
val remainder = targetSum - number;
val remainderResult = howSum(remainder, numbers);
if (remainderResult != null){
return remainderResult + number
}
}
return null
}
I have following code which is working in Objective-C:
NSScanner *scanner ;
for(int i = 0; i < [expression count]; i = i + 2)
{
scanner = [NSScanner scannerWithString:[expression objectAtIndex:i]];
BOOL isNumeric = [scanner scanInteger:NULL] && [scanner isAtEnd];
if(!isNumeric)
return false;
}
return true;
I need equivalent code in Swift 4. I have tried different things but couldn't work it out. The requirement is to check whether the elements of array are number or not.
To check if an object is a number (Int in your case), you could do two things:
Type check via is or as?
This only checks the type and not the content
let isNumberType = "1" is Int
print(isNumberType) //false because "1" is of type String
Creating an Int via it's initializer
This returns an Int? because it can fail so further check != nil
let something = "1"
let isNumber = Int(something) != nil
print(isNumber) //true because "1" can be made into an Int
NOTE: As per your example, you're checking only even elements, hence we will use stride(from:to:by:)
Solution #1:
Assuming you have an array of Strings, we can use the Int initializer to check if the string element can be a number, like so:
func check(expression: [String]) -> Bool {
for idx in stride(from: 0, to: expression.count, by: 2) {
let isNumeric = Int(expression[idx]) != nil
if isNumeric == false {
return false
}
}
return true
}
check(expression: ["1", "A", "2", "B", "3", "C"]) //true
check(expression: ["1", "A", "2", "B", "E", "C"]) //false
Solution #2:
Assuming your array is of type [Any] and you want to type check the alternate elements to be Int then use is, like so:
func check(expression: [Any]) -> Bool {
for idx in stride(from: 0, to: expression.count, by: 2) {
let isNumeric = expression[idx] is Int
if isNumeric == false {
return false
}
}
return true
}
check(expression: [1, "A", 2, "B", 3, "C"]) //true
check(expression: [1, "A", 2, "B", "3", "C"]) //false
The thing with [Any] is that it's elements can't be fed directly to the Int's initializer without bringing it into an acceptable type.
So in this example, for simplicity sake, we are just checking if the object is exactly of type Int or not.
Therefore, I doubt this one suits your requirement.
Try this:
var str = "1234456";
let scanner = Scanner(string: str);
let isNumeric = scanner.scanInt(nil) && scanner.isAtEnd
if !isNumeric {
print("not numeric")
} else {
print("is numeric")
}
Overall if you just want to check the given string is an parsable Integer, I recommend you try :
var expressions = ["1234456","abcd"];
for str in expressions {
if let isNumeric = Int(str) {
print("is numeric")
} else {
print("not numeric")
}
}
Is there a function built into Java that capitalizes the first character of each word in a String, and does not affect the others?
Examples:
jon skeet -> Jon Skeet
miles o'Brien -> Miles O'Brien (B remains capital, this rules out Title Case)
old mcdonald -> Old Mcdonald*
*(Old McDonald would be find too, but I don't expect it to be THAT smart.)
A quick look at the Java String Documentation reveals only toUpperCase() and toLowerCase(), which of course do not provide the desired behavior. Naturally, Google results are dominated by those two functions. It seems like a wheel that must have been invented already, so it couldn't hurt to ask so I can use it in the future.
WordUtils.capitalize(str) (from apache commons-text)
(Note: if you need "fOO BAr" to become "Foo Bar", then use capitalizeFully(..) instead)
If you're only worried about the first letter of the first word being capitalized:
private String capitalize(final String line) {
return Character.toUpperCase(line.charAt(0)) + line.substring(1);
}
The following method converts all the letters into upper/lower case, depending on their position near a space or other special chars.
public static String capitalizeString(String string) {
char[] chars = string.toLowerCase().toCharArray();
boolean found = false;
for (int i = 0; i < chars.length; i++) {
if (!found && Character.isLetter(chars[i])) {
chars[i] = Character.toUpperCase(chars[i]);
found = true;
} else if (Character.isWhitespace(chars[i]) || chars[i]=='.' || chars[i]=='\'') { // You can add other chars here
found = false;
}
}
return String.valueOf(chars);
}
Try this very simple way
example givenString="ram is good boy"
public static String toTitleCase(String givenString) {
String[] arr = givenString.split(" ");
StringBuffer sb = new StringBuffer();
for (int i = 0; i < arr.length; i++) {
sb.append(Character.toUpperCase(arr[i].charAt(0)))
.append(arr[i].substring(1)).append(" ");
}
return sb.toString().trim();
}
Output will be: Ram Is Good Boy
I made a solution in Java 8 that is IMHO more readable.
public String firstLetterCapitalWithSingleSpace(final String words) {
return Stream.of(words.trim().split("\\s"))
.filter(word -> word.length() > 0)
.map(word -> word.substring(0, 1).toUpperCase() + word.substring(1))
.collect(Collectors.joining(" "));
}
The Gist for this solution can be found here: https://gist.github.com/Hylke1982/166a792313c5e2df9d31
String toBeCapped = "i want this sentence capitalized";
String[] tokens = toBeCapped.split("\\s");
toBeCapped = "";
for(int i = 0; i < tokens.length; i++){
char capLetter = Character.toUpperCase(tokens[i].charAt(0));
toBeCapped += " " + capLetter + tokens[i].substring(1);
}
toBeCapped = toBeCapped.trim();
I've written a small Class to capitalize all the words in a String.
Optional multiple delimiters, each one with its behavior (capitalize before, after, or both, to handle cases like O'Brian);
Optional Locale;
Don't breaks with Surrogate Pairs.
LIVE DEMO
Output:
====================================
SIMPLE USAGE
====================================
Source: cApItAlIzE this string after WHITE SPACES
Output: Capitalize This String After White Spaces
====================================
SINGLE CUSTOM-DELIMITER USAGE
====================================
Source: capitalize this string ONLY before'and''after'''APEX
Output: Capitalize this string only beforE'AnD''AfteR'''Apex
====================================
MULTIPLE CUSTOM-DELIMITER USAGE
====================================
Source: capitalize this string AFTER SPACES, BEFORE'APEX, and #AFTER AND BEFORE# NUMBER SIGN (#)
Output: Capitalize This String After Spaces, BeforE'apex, And #After And BeforE# Number Sign (#)
====================================
SIMPLE USAGE WITH CUSTOM LOCALE
====================================
Source: Uniforming the first and last vowels (different kind of 'i's) of the Turkish word D[İ]YARBAK[I]R (DİYARBAKIR)
Output: Uniforming The First And Last Vowels (different Kind Of 'i's) Of The Turkish Word D[i]yarbak[i]r (diyarbakir)
====================================
SIMPLE USAGE WITH A SURROGATE PAIR
====================================
Source: ab 𐐂c de à
Output: Ab 𐐪c De À
Note: first letter will always be capitalized (edit the source if you don't want that).
Please share your comments and help me to found bugs or to improve the code...
Code:
import java.util.ArrayList;
import java.util.Date;
import java.util.List;
import java.util.Locale;
public class WordsCapitalizer {
public static String capitalizeEveryWord(String source) {
return capitalizeEveryWord(source,null,null);
}
public static String capitalizeEveryWord(String source, Locale locale) {
return capitalizeEveryWord(source,null,locale);
}
public static String capitalizeEveryWord(String source, List<Delimiter> delimiters, Locale locale) {
char[] chars;
if (delimiters == null || delimiters.size() == 0)
delimiters = getDefaultDelimiters();
// If Locale specified, i18n toLowerCase is executed, to handle specific behaviors (eg. Turkish dotted and dotless 'i')
if (locale!=null)
chars = source.toLowerCase(locale).toCharArray();
else
chars = source.toLowerCase().toCharArray();
// First charachter ALWAYS capitalized, if it is a Letter.
if (chars.length>0 && Character.isLetter(chars[0]) && !isSurrogate(chars[0])){
chars[0] = Character.toUpperCase(chars[0]);
}
for (int i = 0; i < chars.length; i++) {
if (!isSurrogate(chars[i]) && !Character.isLetter(chars[i])) {
// Current char is not a Letter; gonna check if it is a delimitrer.
for (Delimiter delimiter : delimiters){
if (delimiter.getDelimiter()==chars[i]){
// Delimiter found, applying rules...
if (delimiter.capitalizeBefore() && i>0
&& Character.isLetter(chars[i-1]) && !isSurrogate(chars[i-1]))
{ // previous character is a Letter and I have to capitalize it
chars[i-1] = Character.toUpperCase(chars[i-1]);
}
if (delimiter.capitalizeAfter() && i<chars.length-1
&& Character.isLetter(chars[i+1]) && !isSurrogate(chars[i+1]))
{ // next character is a Letter and I have to capitalize it
chars[i+1] = Character.toUpperCase(chars[i+1]);
}
break;
}
}
}
}
return String.valueOf(chars);
}
private static boolean isSurrogate(char chr){
// Check if the current character is part of an UTF-16 Surrogate Pair.
// Note: not validating the pair, just used to bypass (any found part of) it.
return (Character.isHighSurrogate(chr) || Character.isLowSurrogate(chr));
}
private static List<Delimiter> getDefaultDelimiters(){
// If no delimiter specified, "Capitalize after space" rule is set by default.
List<Delimiter> delimiters = new ArrayList<Delimiter>();
delimiters.add(new Delimiter(Behavior.CAPITALIZE_AFTER_MARKER, ' '));
return delimiters;
}
public static class Delimiter {
private Behavior behavior;
private char delimiter;
public Delimiter(Behavior behavior, char delimiter) {
super();
this.behavior = behavior;
this.delimiter = delimiter;
}
public boolean capitalizeBefore(){
return (behavior.equals(Behavior.CAPITALIZE_BEFORE_MARKER)
|| behavior.equals(Behavior.CAPITALIZE_BEFORE_AND_AFTER_MARKER));
}
public boolean capitalizeAfter(){
return (behavior.equals(Behavior.CAPITALIZE_AFTER_MARKER)
|| behavior.equals(Behavior.CAPITALIZE_BEFORE_AND_AFTER_MARKER));
}
public char getDelimiter() {
return delimiter;
}
}
public static enum Behavior {
CAPITALIZE_AFTER_MARKER(0),
CAPITALIZE_BEFORE_MARKER(1),
CAPITALIZE_BEFORE_AND_AFTER_MARKER(2);
private int value;
private Behavior(int value) {
this.value = value;
}
public int getValue() {
return value;
}
}
Using org.apache.commons.lang.StringUtils makes it very simple.
capitalizeStr = StringUtils.capitalize(str);
From Java 9+
you can use String::replaceAll like this :
public static void upperCaseAllFirstCharacter(String text) {
String regex = "\\b(.)(.*?)\\b";
String result = Pattern.compile(regex).matcher(text).replaceAll(
matche -> matche.group(1).toUpperCase() + matche.group(2)
);
System.out.println(result);
}
Example :
upperCaseAllFirstCharacter("hello this is Just a test");
Outputs
Hello This Is Just A Test
With this simple code:
String example="hello";
example=example.substring(0,1).toUpperCase()+example.substring(1, example.length());
System.out.println(example);
Result: Hello
I'm using the following function. I think it is faster in performance.
public static String capitalize(String text){
String c = (text != null)? text.trim() : "";
String[] words = c.split(" ");
String result = "";
for(String w : words){
result += (w.length() > 1? w.substring(0, 1).toUpperCase(Locale.US) + w.substring(1, w.length()).toLowerCase(Locale.US) : w) + " ";
}
return result.trim();
}
Use the Split method to split your string into words, then use the built in string functions to capitalize each word, then append together.
Pseudo-code (ish)
string = "the sentence you want to apply caps to";
words = string.split(" ")
string = ""
for(String w: words)
//This line is an easy way to capitalize a word
word = word.toUpperCase().replace(word.substring(1), word.substring(1).toLowerCase())
string += word
In the end string looks something like
"The Sentence You Want To Apply Caps To"
This might be useful if you need to capitalize titles. It capitalizes each substring delimited by " ", except for specified strings such as "a" or "the". I haven't ran it yet because it's late, should be fine though. Uses Apache Commons StringUtils.join() at one point. You can substitute it with a simple loop if you wish.
private static String capitalize(String string) {
if (string == null) return null;
String[] wordArray = string.split(" "); // Split string to analyze word by word.
int i = 0;
lowercase:
for (String word : wordArray) {
if (word != wordArray[0]) { // First word always in capital
String [] lowercaseWords = {"a", "an", "as", "and", "although", "at", "because", "but", "by", "for", "in", "nor", "of", "on", "or", "so", "the", "to", "up", "yet"};
for (String word2 : lowercaseWords) {
if (word.equals(word2)) {
wordArray[i] = word;
i++;
continue lowercase;
}
}
}
char[] characterArray = word.toCharArray();
characterArray[0] = Character.toTitleCase(characterArray[0]);
wordArray[i] = new String(characterArray);
i++;
}
return StringUtils.join(wordArray, " "); // Re-join string
}
public static String toTitleCase(String word){
return Character.toUpperCase(word.charAt(0)) + word.substring(1);
}
public static void main(String[] args){
String phrase = "this is to be title cased";
String[] splitPhrase = phrase.split(" ");
String result = "";
for(String word: splitPhrase){
result += toTitleCase(word) + " ";
}
System.out.println(result.trim());
}
1. Java 8 Streams
public static String capitalizeAll(String str) {
if (str == null || str.isEmpty()) {
return str;
}
return Arrays.stream(str.split("\\s+"))
.map(t -> t.substring(0, 1).toUpperCase() + t.substring(1))
.collect(Collectors.joining(" "));
}
Examples:
System.out.println(capitalizeAll("jon skeet")); // Jon Skeet
System.out.println(capitalizeAll("miles o'Brien")); // Miles O'Brien
System.out.println(capitalizeAll("old mcdonald")); // Old Mcdonald
System.out.println(capitalizeAll(null)); // null
For foo bAR to Foo Bar, replace the map() method with the following:
.map(t -> t.substring(0, 1).toUpperCase() + t.substring(1).toLowerCase())
2. String.replaceAll() (Java 9+)
ublic static String capitalizeAll(String str) {
if (str == null || str.isEmpty()) {
return str;
}
return Pattern.compile("\\b(.)(.*?)\\b")
.matcher(str)
.replaceAll(match -> match.group(1).toUpperCase() + match.group(2));
}
Examples:
System.out.println(capitalizeAll("12 ways to learn java")); // 12 Ways To Learn Java
System.out.println(capitalizeAll("i am atta")); // I Am Atta
System.out.println(capitalizeAll(null)); // null
3. Apache Commons Text
System.out.println(WordUtils.capitalize("love is everywhere")); // Love Is Everywhere
System.out.println(WordUtils.capitalize("sky, sky, blue sky!")); // Sky, Sky, Blue Sky!
System.out.println(WordUtils.capitalize(null)); // null
For titlecase:
System.out.println(WordUtils.capitalizeFully("fOO bAR")); // Foo Bar
System.out.println(WordUtils.capitalizeFully("sKy is BLUE!")); // Sky Is Blue!
For details, checkout this tutorial.
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the sentence : ");
try
{
String str = br.readLine();
char[] str1 = new char[str.length()];
for(int i=0; i<str.length(); i++)
{
str1[i] = Character.toLowerCase(str.charAt(i));
}
str1[0] = Character.toUpperCase(str1[0]);
for(int i=0;i<str.length();i++)
{
if(str1[i] == ' ')
{
str1[i+1] = Character.toUpperCase(str1[i+1]);
}
System.out.print(str1[i]);
}
}
catch(Exception e)
{
System.err.println("Error: " + e.getMessage());
}
I decided to add one more solution for capitalizing words in a string:
words are defined here as adjacent letter-or-digit characters;
surrogate pairs are provided as well;
the code has been optimized for performance; and
it is still compact.
Function:
public static String capitalize(String string) {
final int sl = string.length();
final StringBuilder sb = new StringBuilder(sl);
boolean lod = false;
for(int s = 0; s < sl; s++) {
final int cp = string.codePointAt(s);
sb.appendCodePoint(lod ? Character.toLowerCase(cp) : Character.toUpperCase(cp));
lod = Character.isLetterOrDigit(cp);
if(!Character.isBmpCodePoint(cp)) s++;
}
return sb.toString();
}
Example call:
System.out.println(capitalize("An à la carte StRiNg. Surrogate pairs: 𐐪𐐪."));
Result:
An À La Carte String. Surrogate Pairs: 𐐂𐐪.
Use:
String text = "jon skeet, miles o'brien, old mcdonald";
Pattern pattern = Pattern.compile("\\b([a-z])([\\w]*)");
Matcher matcher = pattern.matcher(text);
StringBuffer buffer = new StringBuffer();
while (matcher.find()) {
matcher.appendReplacement(buffer, matcher.group(1).toUpperCase() + matcher.group(2));
}
String capitalized = matcher.appendTail(buffer).toString();
System.out.println(capitalized);
There are many way to convert the first letter of the first word being capitalized. I have an idea. It's very simple:
public String capitalize(String str){
/* The first thing we do is remove whitespace from string */
String c = str.replaceAll("\\s+", " ");
String s = c.trim();
String l = "";
for(int i = 0; i < s.length(); i++){
if(i == 0){ /* Uppercase the first letter in strings */
l += s.toUpperCase().charAt(i);
i++; /* To i = i + 1 because we don't need to add
value i = 0 into string l */
}
l += s.charAt(i);
if(s.charAt(i) == 32){ /* If we meet whitespace (32 in ASCII Code is whitespace) */
l += s.toUpperCase().charAt(i+1); /* Uppercase the letter after whitespace */
i++; /* Yo i = i + 1 because we don't need to add
value whitespace into string l */
}
}
return l;
}
package com.test;
/**
* #author Prasanth Pillai
* #date 01-Feb-2012
* #description : Below is the test class details
*
* inputs a String from a user. Expect the String to contain spaces and alphanumeric characters only.
* capitalizes all first letters of the words in the given String.
* preserves all other characters (including spaces) in the String.
* displays the result to the user.
*
* Approach : I have followed a simple approach. However there are many string utilities available
* for the same purpose. Example : WordUtils.capitalize(str) (from apache commons-lang)
*
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Test {
public static void main(String[] args) throws IOException{
System.out.println("Input String :\n");
InputStreamReader converter = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(converter);
String inputString = in.readLine();
int length = inputString.length();
StringBuffer newStr = new StringBuffer(0);
int i = 0;
int k = 0;
/* This is a simple approach
* step 1: scan through the input string
* step 2: capitalize the first letter of each word in string
* The integer k, is used as a value to determine whether the
* letter is the first letter in each word in the string.
*/
while( i < length){
if (Character.isLetter(inputString.charAt(i))){
if ( k == 0){
newStr = newStr.append(Character.toUpperCase(inputString.charAt(i)));
k = 2;
}//this else loop is to avoid repeatation of the first letter in output string
else {
newStr = newStr.append(inputString.charAt(i));
}
} // for the letters which are not first letter, simply append to the output string.
else {
newStr = newStr.append(inputString.charAt(i));
k=0;
}
i+=1;
}
System.out.println("new String ->"+newStr);
}
}
Here is a simple function
public static String capEachWord(String source){
String result = "";
String[] splitString = source.split(" ");
for(String target : splitString){
result += Character.toUpperCase(target.charAt(0))
+ target.substring(1) + " ";
}
return result.trim();
}
This is just another way of doing it:
private String capitalize(String line)
{
StringTokenizer token =new StringTokenizer(line);
String CapLine="";
while(token.hasMoreTokens())
{
String tok = token.nextToken().toString();
CapLine += Character.toUpperCase(tok.charAt(0))+ tok.substring(1)+" ";
}
return CapLine.substring(0,CapLine.length()-1);
}
Reusable method for intiCap:
public class YarlagaddaSireeshTest{
public static void main(String[] args) {
String FinalStringIs = "";
String testNames = "sireesh yarlagadda test";
String[] name = testNames.split("\\s");
for(String nameIs :name){
FinalStringIs += getIntiCapString(nameIs) + ",";
}
System.out.println("Final Result "+ FinalStringIs);
}
public static String getIntiCapString(String param) {
if(param != null && param.length()>0){
char[] charArray = param.toCharArray();
charArray[0] = Character.toUpperCase(charArray[0]);
return new String(charArray);
}
else {
return "";
}
}
}
Here is my solution.
I ran across this problem tonight and decided to search it. I found an answer by Neelam Singh that was almost there, so I decided to fix the issue (broke on empty strings) and caused a system crash.
The method you are looking for is named capString(String s) below.
It turns "It's only 5am here" into "It's Only 5am Here".
The code is pretty well commented, so enjoy.
package com.lincolnwdaniel.interactivestory.model;
public class StringS {
/**
* #param s is a string of any length, ideally only one word
* #return a capitalized string.
* only the first letter of the string is made to uppercase
*/
public static String capSingleWord(String s) {
if(s.isEmpty() || s.length()<2) {
return Character.toUpperCase(s.charAt(0))+"";
}
else {
return Character.toUpperCase(s.charAt(0)) + s.substring(1);
}
}
/**
*
* #param s is a string of any length
* #return a title cased string.
* All first letter of each word is made to uppercase
*/
public static String capString(String s) {
// Check if the string is empty, if it is, return it immediately
if(s.isEmpty()){
return s;
}
// Split string on space and create array of words
String[] arr = s.split(" ");
// Create a string buffer to hold the new capitalized string
StringBuffer sb = new StringBuffer();
// Check if the array is empty (would be caused by the passage of s as an empty string [i.g "" or " "],
// If it is, return the original string immediately
if( arr.length < 1 ){
return s;
}
for (int i = 0; i < arr.length; i++) {
sb.append(Character.toUpperCase(arr[i].charAt(0)))
.append(arr[i].substring(1)).append(" ");
}
return sb.toString().trim();
}
}
Here we go for perfect first char capitalization of word
public static void main(String[] args) {
String input ="my name is ranjan";
String[] inputArr = input.split(" ");
for(String word : inputArr) {
System.out.println(word.substring(0, 1).toUpperCase()+word.substring(1,word.length()));
}
}
}
//Output : My Name Is Ranjan
For those of you using Velocity in your MVC, you can use the capitalizeFirstLetter() method from the StringUtils class.
String s="hi dude i want apple";
s = s.replaceAll("\\s+"," ");
String[] split = s.split(" ");
s="";
for (int i = 0; i < split.length; i++) {
split[i]=Character.toUpperCase(split[i].charAt(0))+split[i].substring(1);
s+=split[i]+" ";
System.out.println(split[i]);
}
System.out.println(s);
package corejava.string.intern;
import java.io.DataInputStream;
import java.util.ArrayList;
/*
* wap to accept only 3 sentences and convert first character of each word into upper case
*/
public class Accept3Lines_FirstCharUppercase {
static String line;
static String words[];
static ArrayList<String> list=new ArrayList<String>();
/**
* #param args
*/
public static void main(String[] args) throws java.lang.Exception{
DataInputStream read=new DataInputStream(System.in);
System.out.println("Enter only three sentences");
int i=0;
while((line=read.readLine())!=null){
method(line); //main logic of the code
if((i++)==2){
break;
}
}
display();
System.out.println("\n End of the program");
}
/*
* this will display all the elements in an array
*/
public static void display(){
for(String display:list){
System.out.println(display);
}
}
/*
* this divide the line of string into words
* and first char of the each word is converted to upper case
* and to an array list
*/
public static void method(String lineParam){
words=line.split("\\s");
for(String s:words){
String result=s.substring(0,1).toUpperCase()+s.substring(1);
list.add(result);
}
}
}
If you prefer Guava...
String myString = ...;
String capWords = Joiner.on(' ').join(Iterables.transform(Splitter.on(' ').omitEmptyStrings().split(myString), new Function<String, String>() {
public String apply(String input) {
return Character.toUpperCase(input.charAt(0)) + input.substring(1);
}
}));
String toUpperCaseFirstLetterOnly(String str) {
String[] words = str.split(" ");
StringBuilder ret = new StringBuilder();
for(int i = 0; i < words.length; i++) {
ret.append(Character.toUpperCase(words[i].charAt(0)));
ret.append(words[i].substring(1));
if(i < words.length - 1) {
ret.append(' ');
}
}
return ret.toString();
}
I am trying replicate some Objective C cocoa in Swift. All is good until I come across the following:
// Set a new type and creator:
unsigned long type = 'TEXT';
unsigned long creator = 'pdos';
How can I create Int64s (or the correct Swift equivalent) from single quote character literals like this?
Types:
public typealias AEKeyword = FourCharCode
public typealias OSType = FourCharCode
public typealias FourCharCode = UInt32
I'm using this in my Cocoa Scripting apps, it considers characters > 0x80 correctly
func OSTypeFrom(string : String) -> UInt {
var result : UInt = 0
if let data = string.dataUsingEncoding(NSMacOSRomanStringEncoding) {
let bytes = UnsafePointer<UInt8>(data.bytes)
for i in 0..<data.length {
result = result << 8 + UInt(bytes[i])
}
}
return result
}
Edit:
Alternatively
func fourCharCodeFrom(string : String) -> FourCharCode
{
assert(string.count == 4, "String length must be 4")
var result : FourCharCode = 0
for char in string.utf16 {
result = (result << 8) + FourCharCode(char)
}
return result
}
or still swiftier
func fourCharCode(from string : String) -> FourCharCode
{
return string.utf16.reduce(0, {$0 << 8 + FourCharCode($1)})
}
I found the following typealiases from the Swift API:
typealias FourCharCode = UInt32
typealias OSType = FourCharCode
And the following functions:
func NSFileTypeForHFSTypeCode(hfsFileTypeCode: OSType) -> String!
func NSHFSTypeCodeFromFileType(fileTypeString: String!) -> OSType
This should allow me to create the equivalent code:
let type : UInt32 = UInt32(NSHFSTypeCodeFromFileType("TEXT"))
let creator : UInt32 = UInt32(NSHFSTypeCodeFromFileType("pdos"))
But those 4-character strings doesn't work and return 0.
If you wrap each string in ' single quotes ' and call the same functions, you will get the correct return values:
let type : UInt32 = UInt32(NSHFSTypeCodeFromFileType("'TEXT'"))
let creator : UInt32 = UInt32(NSHFSTypeCodeFromFileType("'pdos'"))
Adopt the ExpressibleByStringLiteral protocol to use four-character string literals directly:
extension FourCharCode: ExpressibleByStringLiteral {
public init(stringLiteral value: StringLiteralType) {
if let data = value.data(using: .macOSRoman), data.count == 4 {
self = data.reduce(0, {$0 << 8 + Self($1)})
} else {
self = 0
}
}
}
Now you can just pass a string literal as the FourCharCode / OSType / UInt32 parameter:
let record = NSAppleEventDescriptor.record()
record.setDescriptor(NSAppleEventDescriptor(boolean: true), forKeyword: "test")
In Swift 4 or later, I use this code - if the string is not 4 characters in size, it will return an OSType(0):
extension String {
public func osType() -> OSType {
var result:UInt = 0
if let data = self.data(using: .macOSRoman), data.count == 4
{
data.withUnsafeBytes { (ptr:UnsafePointer<UInt8>) in
for i in 0..<data.count {
result = result << 8 + UInt(ptr[i])
}
}
}
return OSType(result)
}
}
let type = "APPL".osType() // 1095782476
// check if this is OK in a playground
let hexStr = String(format: "0x%lx", type) // 0x4150504c -> "APPL" in ASCII
Swift 5 Update:
extension String {
func osType() -> OSType {
return OSType(
data(using: .macOSRoman)?
.withUnsafeBytes {
$0.reduce(into: UInt(0)) { $0 = $0 << 8 + UInt($1) }
} ?? 0
)
}
}
Here's a simple function
func mbcc(foo: String) -> Int
{
let chars = foo.utf8
var result: Int = 0
for aChar in chars
{
result = result << 8 + Int(aChar)
}
return result
}
let a = mbcc("TEXT")
print(String(format: "0x%lx", a)) // Prints 0x54455854
It will work for strings that will fit in an Int. Once they get longer it starts losing digits from the top.
If you use
result = result * 256 + Int(aChar)
you should get a crash when the string gets too big instead.
Using NSHFSTypeCodeFromFileType does work, but only for 4-character strings wrapped with single quotes, aka 6-character strings. It returns 0 for unquoted 4-character strings.
So wrap your 4-character string in ' ' before passing it to the function:
extension FourCharCode: ExpressibleByStringLiteral {
public init(stringLiteral value: StringLiteralType) {
switch (value.count, value.first, value.last) {
case (6, "'", "'"):
self = NSHFSTypeCodeFromFileType(value)
case (4, _, _):
self = NSHFSTypeCodeFromFileType("'\(value)'")
default:
self = 0
}
}
}
Using the above extension, you can use 4-character or single-quoted 6-character string literals:
let record = NSAppleEventDescriptor.record()
record.setDescriptor(NSAppleEventDescriptor(boolean: true), forKeyword: "4444")
record.setDescriptor(NSAppleEventDescriptor(boolean: true), forKeyword: "'6666'")
It would be even better to limit the string literal to 4-character strings at compile time. That does not seem to currently be possible, but is being discussed for Swift here:
Allow for Compile-Time Checked Intervals for Parameters Expecting Literal Values