Question

我试图在java中获取字符串输入。我的输入应该是这样的

3
1,1,bugs@bunny.com,123 Sesame St.,New York,NY,10011,12345689010
1,2,bugs@bunny.com,123 Sesame St.,New York,NY,10011,12345689010
1,3,bugs@bunny.com,123 Sesame St.,New York,NY,10011,12345689010

所以，我试过这个

Scanner in = new Scanner(System.in);
        int TotalNumber = in.nextInt();
        String[] Data = new String[TotalNumber];
        for (int Counter = 0; Counter < TotalNumber; Counter++) {

            Data[Counter] = in.next();

        }

        in.close();

        for (int counter = 0; counter < Data.length; counter++) {


            System.out.println(Data[counter]);

        }

我的输出显示了这个

1,1,bugs@bunny.com,123
Sesame
St.,New

我的问题是什么？如何正确输入字符串？

更新

我在Scanner issue when using nextLine after nextXXX

找到了我的解决方案

Answer 1

next()在空白处打破。相反，您应该使用nextLine()将整行输入到字符串中：

int TotalNumber = in.nextInt();
String[] Data = new String[TotalNumber];
for (int Counter = 0; Counter < TotalNumber; Counter++) {
    Data[Counter] = in.nextLine();
}

Answer 2

尝试使用Data[Counter] = in.nextLine();

Answer 3

怎么样：

import java.util.Scanner;
import java.lang.String;

public class Test
{
    public static void main(String[] args)
    {
        char[] sArray;

        Scanner scan = new Scanner(System.in);

        System.out.print("Enter a Palindrome : ");
        String s = scan.nextLine();
        s = s.replaceAll("\\s+", "");

        sArray = new char[s.length()];

        for(int i = 0; i < s.length(); i++)
        {
            sArray[i] = s.charAt(i);
            System.out.print(sArray[i]);
        }
    }
}

Answer 4

试试这个（Mureinik修改过的代码）..

int TotalNumber = in.nextInt();
in.nextLine();
String[] Data = new String[TotalNumber];
for (int Counter = 0; Counter < TotalNumber; Counter++) {
    Data[Counter] = in.nextLine();
}

您在获取nextLine()后需要int，因为您在执行int后会按Enter键，并且nextLine()中的Data[0]会读取该输入。

Java从控制台获取字符串输入

4 个答案: