Java String

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An immutable, iterable, non-primitive data type, consisting of characters, implementing CharSequence, Serializable, Comparable<String> interfaces.

String name = "Fossery";

Use variable values to create Strings

Concatenation

Using + operator:

String hello = "Hello " + name + "!";

Using .concat(String str) method:

String hello2 = "Hello ".concat(name).concat("!"); // concat() can only have one argument

Using StringBuilder or StringBuffer (the .append() and .toString() methods of both classes give the same result):

StringBuilder sb = new StringBuilder();
sb.append("Hello ");
sb.append(name);
sb.append("!");
String hello3 = sb.toString();

// or

StringBuilder sb = new StringBuilder();
sb.append("Hello ").append(name).append("!"); // append() can only have one argument
String hello3 = sb.toString();

String literal

Using String.format(String pattern, Object... arguments):

String hello4 = String.format("Hello %s!", name);

Using MessageFormat.format(String pattern, Object... arguments):

import java.text.MessageFormat;

String hello5 = MessageFormat.format("Hello {0}!", name);

Get length of String

Using .length() method:

// name = "Fossery"
int nameLength = name.length(); // 7

Compare two Strings

Using .equals(Object obj) method - return true if the two strings match, false if not:

boolean doTheyMatch = name.equals("Fossery"); // true

Using .compareTo(String str) method - compare lexicografically, return negative integer if original String precedes argument String, 0 if they match, positive integer if argument String precedes original String:

int nameComparison = name.compareTo("Fossery"); // 0

Important! == operator shouldn't be used for comparing Strings (or any objects of non-primitive type) in most cases, because it compares the references of the two Strings, so "Fossery" == "Fossery" will return false, because those are two different String objects, regardless of all their characters matching.

Ignoring case

Using .equalsIgnoreCase(String str) or .compareToIgnoreCase(String str) - work similarly to .equals(Object obj) and .compareTo(String str).

Get character at a specific index

Using .charAt(int index) method:

char initialLetter = name.charAt(0); // 'F'

Get a substring

Using .substring(int beginIndex, int endIndex) method - endIndex is optional, selects all characters from beginIndex to the end of the String if not provided:

String foss = name.substring(0, 4); // "Foss"
String ery = name.substring(4); // "Foss"

Search a character in a String

Using .indexOf(char c) method - return index of first occurrence, or -1 if not found (optional fromIndex argument can be provided to start search in that position):

int firstS = name.indexOf('s'); // 2

Using .lastIndexOf(char c) method - return index of last occurrence, or -1 if not found (optional fromIndex argument can be provided to start search in that position):

int lastS = name.lastIndexOf('s'); // 3

Search a substring in a String

Using .indexOf(String c) or .lastIndexOf(String c): use as shown in "Search a character in a String", but pass String instead of char

Using .contains(CharSequence s) method - return boolean:

boolean isHeFoss = name.contains("Foss"); // true

Using .startsWith(String s) method - return boolean (true only if substring is at the start of String):

boolean startsWithFoss = name.startsWith("Foss"); // true

Using .endsWith(String s) method - return boolean (true only if substring is at the end of String):

boolean endsWithFoss = name.endsWith("Foss"); // false

Convert char to String

Using String.valueOf(char character) - most efficient method:

// initialLetter = 'F'
String initialLetterStr = String.valueOf(initialLetter); // "F"

Using Character.toString():

String initialLetterStr2 = new Character(initialLetter).toString(); // "F"

// or

String initialLetterStr3 = Character.toString(initialLetter); // "F" - makes internal call to String.valueOf()

Using + operator (not recommended!):

String initialLetterStr3 = "" + initialLetter; // "F"

Iterate through characters of String

Using for loop - most optimal:

for (int i = 0; i < name.length(); i++) {
  System.out.println(name.charAt(i));
}

// performance-optimized for loop (evaluate name.length() only once):

for (int i = 0, n = name.length(); i < n; i++) {
  System.out.println(name.charAt(i));
}

Converting String to char array and using for-each loop - more readable, but less optimal:

for (char letter : name.toCharArray()) {
  System.out.println(letter);
}

Using .chars() or .codePoints(), with .forEachOrdered() - from Java 8:

name.chars().forEachOrdered(i -> System.out.print((char)i));
name.codePoints().forEachOrdered(i -> System.out.print((char)i));

Join multiple Strings into a single String

Using String.join(CharSequence delimiter, CharSequence... elements):

String shoppingList = String.join(", ", "apple", "bread", "milk"); // "apple, bread, milk"

Using StringJoiner:

import java.util.StringJoiner;

String shoppingList2 = new StringJoiner(", ").add("apple").add("bread").add("milk").toString(); // "apple, bread, milk"

Join elements of String array, List and other Iterables

Using String.join(CharSequence delimiter, Iterable elements):

String[] shoppingListArr = new String[]{"apple", "bread", "milk"};
String shoppingListStr = String.join(", ", shoppingList);// "apple, bread, milk"

Using stream:

import java.util.Arrays;
import java.util.ArrayList;
import java.util.stream.Collectors;

// for Collections

ArrayList<String> shoppingArrayList = new ArrayList<>(Arrays.asList("apple", "bread", "milk"));
String shoppingList3 = shoppingArrayList .stream().collect(Collectors.joining(", ")); // "apple, bread, milk"

// for arrays

String[] shoppingArray = new String[]{"apple", "bread", "milk"};
String shoppingList4 = Arrays.stream(shoppingArray).collect(Collectors.joining(", ")); // "apple, bread, milk"

Split String

Into array

Using .split(String delimiter) method:

String[] shoppingListSplitted = shoppingList.split(", "); // ["apple", "bread", "milk"]

Using Pattern (from Java 11):

import java.util.regex.Pattern;

String[] shoppingListSplitted2 = Pattern.compile(", ")
      .splitAsStream(shoppingList)
      .toArray(String[]::new); // without argument, toArray() returns Object[] which cannot be casted to String[]

Into List

Using Pattern (from Java 8):

import java.util.List;
import java.util.regex.Pattern;
import java.util.stream.Collectors;

List<String> shoppingListSplitted3 = Pattern.compile(", ")
      .splitAsStream(shoppingList)
      .collect(Collectors.toList());

Into Set

Using Pattern (from Java 8):

import java.util.Set;
import java.util.regex.Pattern;
import java.util.stream.Collectors;

Set<String> shoppingListSplitted4 = Pattern.compile(", ")
    .splitAsStream(shoppingList)
    .collect(Collectors.toSet());

Convert String into char array

Using .toCharArray() method:

char[] nameLetters = name.toCharArray(); // ['F', 'o', 's', 's', 'e', 'r', 'y']