Home > Java > Java Basics > body text

How to convert string into character array in java

angryTom
Release: 2019-11-14 15:07:59
Original
61529 people have browsed it

How to convert string into character array in java

How to convert a string into a character array in java

1. Convert a string into an array Method:

There is a String.split() method in the java.lang package. In java, split() is usually used to split a string. What is returned is an array.

Special escape characters must be added with "\\" ("." and "|" are both escape characters)

When programming in Java language, use the "Password Field" jPasswordField component If you want to get the password value, you need to use the getPassword() method of the component. The getPassword() method of jPasswordField returns an array of char type. We often need to convert this array to String type in order to perform tasks such as password matching or password matching. Assignment and other operations. At this time, you need to convert the char type array. Of course, we often encounter situations where the String type is converted to a char array.

2. Use the String.toCharArray() method to convert the string into a character array

public class Test {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner input = new Scanner(System.in);
        String str = input.next();
        char ss[] = str.toCharArray(); //利用toCharArray方法转换
        for (int i = 0; i < ss.length; i++) {
            System.out.println(ss[i]);
        }
    }

}
Copy after login

php Chinese website, a large number of free Java entry Tutorial , welcome to learn online!

The above is the detailed content of How to convert string into character array in java. For more information, please follow other related articles on the PHP Chinese website!

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact [email protected]
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!