Der Teilzeichenfolge der Java-String-Klasse () Die Methode gibt einen Teil der Zeichenfolge zurück.
Wir übergeben die Zahlenpositionen von beginIndex und endIndex in der Java-Teilzeichenfolgenmethode, wobei beginIndex inklusiv und endIndex exklusiv ist. Mit anderen Worten, der beginIndex beginnt bei 0, während der endIndex bei 1 beginnt.
Es gibt zwei Arten von Teilstring-Methoden in Java-Strings.
Unterschrift
public String substring(int startIndex) // type - 1 and public String substring(int startIndex, int endIndex) // type - 2
Wenn wir endIndex nicht angeben, gibt die Methode alle Zeichen von startIndex zurück.
Parameter
Startindex : Der Startindex ist inklusive
endIndex : Endindex ist exklusiv
Kehrt zurück
angegebene Zeichenfolge
Ausnahmewürfe
StringIndexOutOfBoundsException wird ausgelöst, wenn eine der folgenden Bedingungen erfüllt ist.
- wenn der Startindex ein negativer Wert ist
- Der Endindex ist niedriger als der Startindex.
- Entweder der Start- oder Endindex ist größer als die Gesamtzahl der in der Zeichenfolge vorhandenen Zeichen.
Interner Implementierungsteilstring (int beginIndex)
public String substring(int beginIndex) { if (beginIndex <0) { throw new stringindexoutofboundsexception(beginindex); } int sublen="value.length" - beginindex; if (sublen < 0) stringindexoutofboundsexception(sublen); return (beginindex="=" ? this : string(value, beginindex, sublen); pre> <h3>Internal implementation substring(int beginIndex, int endIndex) </h3> <pre> public String substring(int beginIndex, int endIndex) { if (beginIndex value.length) { throw new StringIndexOutOfBoundsException(endIndex); } int subLen = endIndex - beginIndex; if (subLen <0) { throw new stringindexoutofboundsexception(sublen); } return ((beginindex="=" 0) && (endindex="=" value.length)) ? this : string(value, beginindex, sublen); < pre> <h2>Java String substring() method example</h2> <p> <strong>FileName:</strong> SubstringExample.java</p> <pre> public class SubstringExample{ public static void main(String args[]){ String s1='javatpoint'; System.out.println(s1.substring(2,4));//returns va System.out.println(s1.substring(2));//returns vatpoint }} </pre> <span> Test it Now </span> <p> <strong>Output:</strong> </p> <pre>va vatpoint </pre> <h2>Java String substring() Method Example 2</h2> <p> <strong>FileName:</strong> SubstringExample2.java</p> <pre> public class SubstringExample2 { public static void main(String[] args) { String s1='Javatpoint'; String substr = s1.substring(0); // Starts with 0 and goes to end System.out.println(substr); String substr2 = s1.substring(5,10); // Starts from 5 and goes to 10 System.out.println(substr2); String substr3 = s1.substring(5,15); // Returns Exception } } </pre> <p> <strong>Output:</strong> </p> <pre> Javatpoint point Exception in thread 'main' java.lang.StringIndexOutOfBoundsException: begin 5, end 15, length 10 </pre> <h2>Applications of substring() Method</h2> <p>1) The substring() method can be used to do some prefix or suffix extraction. For example, we can have a list of names, and it is required to filter out names with surname as 'singh'. The following program shows the same.</p> <p> <strong>FileName:</strong> SubstringExample3.java</p> <pre> public class SubstringExample3 { // main method public static void main(String argvs[]) { String str[] = { 'Praveen Kumar', 'Yuvraj Singh', 'Harbhajan Singh', 'Gurjit Singh', 'Virat Kohli', 'Rohit Sharma', 'Sandeep Singh', 'Milkha Singh' }; String surName = 'Singh'; int surNameSize = surName.length(); int size = str.length; for(int j = 0; j <size; j++) { int length="str[j].length();" extracting the surname string substr="str[j].substring(length" - surnamesize); checks whether is equal to 'singh' or not if(substr.equals(surname)) system.out.println(str[j]); } < pre> <p> <strong>Output:</strong> </p> <pre> Yuvraj Singh Harbhajan Singh Gurjit Singh Sandeep Singh Milkha Singh </pre> <p>2) The substring() method can also be used to check whether a string is a palindrome or not.</p> <p> <strong>FileName:</strong> SubstringExample4.java</p> <pre> public class SubstringExample4 { public boolean isPalindrome(String str) { int size = str.length(); // handling the base case if(size == 0 || size == 1) { // an empty string // or a string of only one character // is always a palindrome return true; } String f = str.substring(0, 1); String l = str.substring(size - 1); // comparing first and the last character of the string if(l.equals(f)) { // recursively finding the solution using the substring() method // reducing the number of characters of the by 2 for the next recursion return isPalindrome(str.substring(1, size - 1)); } return false; } // main method public static void main(String argvs[]) { // instantiating the class SubstringExample4 SubstringExample4 obj = new SubstringExample4(); String str[] = { 'madam', 'rock', 'eye', 'noon', 'kill' }; int size = str.length; for(int j = 0; j <size; j++) { if(obj.ispalindrome(str[j])) system.out.println(str[j] + ' is a palindrome.'); } else not < pre> <p> <strong>Output:</strong> </p> <pre> madam is a palindrome. rock is not a palindrome. eye is a palindrome. noon is a palindrome. kill is not a palindrome. </pre> <hr></size;></pre></size;></pre></0)></pre></0)> Testen Sie es jetzt Ausgabe:
va vatpoint
Beispiel 2 der Java String substring()-Methode
Dateiname: SubstringExample2.java
public class SubstringExample2 { public static void main(String[] args) { String s1='Javatpoint'; String substr = s1.substring(0); // Starts with 0 and goes to end System.out.println(substr); String substr2 = s1.substring(5,10); // Starts from 5 and goes to 10 System.out.println(substr2); String substr3 = s1.substring(5,15); // Returns Exception } } Ausgabe:
Javatpoint point Exception in thread 'main' java.lang.StringIndexOutOfBoundsException: begin 5, end 15, length 10
Anwendungen der substring()-Methode
1) Die Methode substring() kann verwendet werden, um Präfixe oder Suffixe zu extrahieren. Beispielsweise können wir eine Liste mit Namen haben und es ist erforderlich, Namen mit dem Nachnamen „Singh“ herauszufiltern. Das folgende Programm zeigt dasselbe.
Dateiname: SubstringExample3.java
public class SubstringExample3 { // main method public static void main(String argvs[]) { String str[] = { 'Praveen Kumar', 'Yuvraj Singh', 'Harbhajan Singh', 'Gurjit Singh', 'Virat Kohli', 'Rohit Sharma', 'Sandeep Singh', 'Milkha Singh' }; String surName = 'Singh'; int surNameSize = surName.length(); int size = str.length; for(int j = 0; j <size; j++) { int length="str[j].length();" extracting the surname string substr="str[j].substring(length" - surnamesize); checks whether is equal to \'singh\' or not if(substr.equals(surname)) system.out.println(str[j]); } < pre> <p> <strong>Output:</strong> </p> <pre> Yuvraj Singh Harbhajan Singh Gurjit Singh Sandeep Singh Milkha Singh </pre> <p>2) The substring() method can also be used to check whether a string is a palindrome or not.</p> <p> <strong>FileName:</strong> SubstringExample4.java</p> <pre> public class SubstringExample4 { public boolean isPalindrome(String str) { int size = str.length(); // handling the base case if(size == 0 || size == 1) { // an empty string // or a string of only one character // is always a palindrome return true; } String f = str.substring(0, 1); String l = str.substring(size - 1); // comparing first and the last character of the string if(l.equals(f)) { // recursively finding the solution using the substring() method // reducing the number of characters of the by 2 for the next recursion return isPalindrome(str.substring(1, size - 1)); } return false; } // main method public static void main(String argvs[]) { // instantiating the class SubstringExample4 SubstringExample4 obj = new SubstringExample4(); String str[] = { 'madam', 'rock', 'eye', 'noon', 'kill' }; int size = str.length; for(int j = 0; j <size; j++) { if(obj.ispalindrome(str[j])) system.out.println(str[j] + \' is a palindrome.\'); } else not < pre> <p> <strong>Output:</strong> </p> <pre> madam is a palindrome. rock is not a palindrome. eye is a palindrome. noon is a palindrome. kill is not a palindrome. </pre> <hr></size;></pre></size;> 2) Mit der Methode substring() kann auch überprüft werden, ob ein String ein Palindrom ist oder nicht.
Dateiname: SubstringExample4.java
public class SubstringExample4 { public boolean isPalindrome(String str) { int size = str.length(); // handling the base case if(size == 0 || size == 1) { // an empty string // or a string of only one character // is always a palindrome return true; } String f = str.substring(0, 1); String l = str.substring(size - 1); // comparing first and the last character of the string if(l.equals(f)) { // recursively finding the solution using the substring() method // reducing the number of characters of the by 2 for the next recursion return isPalindrome(str.substring(1, size - 1)); } return false; } // main method public static void main(String argvs[]) { // instantiating the class SubstringExample4 SubstringExample4 obj = new SubstringExample4(); String str[] = { 'madam', 'rock', 'eye', 'noon', 'kill' }; int size = str.length; for(int j = 0; j <size; j++) { if(obj.ispalindrome(str[j])) system.out.println(str[j] + \\' is a palindrome.\\'); } else not < pre> <p> <strong>Output:</strong> </p> <pre> madam is a palindrome. rock is not a palindrome. eye is a palindrome. noon is a palindrome. kill is not a palindrome. </pre> <hr></size;>