public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int i = scan.nextInt();
        scan.nextLine();
        double d = scan.nextDouble();
        scan.nextLine();
        String s = scan.nextLine();

        System.out.println("String: " + s);
        System.out.println("Double: " + d);
        System.out.println("Int: " + i);
    }

问题出在input.nextInt()方法;它只读取int值。因此，当您继续使用input.nextLine()读取时，您会收到“\n”输入键。所以要跳过这个，你必须添加input.nextLine()。

试试这样吧:

System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();

使用input.next()而不是input.nextLine()，这应该可以解决问题。

修改代码:

public static Scanner input = new Scanner(System.in);

public static void main(String[] args)
{
    System.out.print("Insert a number: ");
    int number = input.nextInt();
    System.out.print("Text1: ");
    String text1 = input.next();
    System.out.print("Text2: ");
    String text2 = input.next();
}

这是因为当您输入一个数字然后按enter时，input.nextInt()只使用该数字，而不是“行结束”。当input. nextline()执行时，它使用仍然在第一个输入缓冲区中的“行尾”。

相反，在input.nextInt()之后立即使用input.nextLine()

如果你想快速扫描输入而不混淆到扫描器类nextLine()方法，请使用自定义输入扫描器。

代码:

class ScanReader {
/**
* @author Nikunj Khokhar
*/
    private byte[] buf = new byte[4 * 1024];
    private int index;
    private BufferedInputStream in;
    private int total;

    public ScanReader(InputStream inputStream) {
        in = new BufferedInputStream(inputStream);
    }

    private int scan() throws IOException {
        if (index >= total) {
            index = 0;
            total = in.read(buf);
            if (total <= 0) return -1;
        }
        return buf[index++];
    }
    public char scanChar(){
        int c=scan();
        while (isWhiteSpace(c))c=scan();
        return (char)c;
    }


    public int scanInt() throws IOException {
        int integer = 0;
        int n = scan();
        while (isWhiteSpace(n)) n = scan();
        int neg = 1;
        if (n == '-') {
            neg = -1;
            n = scan();
        }
        while (!isWhiteSpace(n)) {
            if (n >= '0' && n <= '9') {
                integer *= 10;
                integer += n - '0';
                n = scan();
            }
        }
        return neg * integer;
    }

    public String scanString() throws IOException {
        int c = scan();
        while (isWhiteSpace(c)) c = scan();
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = scan();
        } while (!isWhiteSpace(c));
        return res.toString();
    }

    private boolean isWhiteSpace(int n) {
        if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
        else return false;
    }

    public long scanLong() throws IOException {
        long integer = 0;
        int n = scan();
        while (isWhiteSpace(n)) n = scan();
        int neg = 1;
        if (n == '-') {
            neg = -1;
            n = scan();
        }
        while (!isWhiteSpace(n)) {
            if (n >= '0' && n <= '9') {
                integer *= 10;
                integer += n - '0';
                n = scan();
            }
        }
        return neg * integer;
    }

    public void scanLong(long[] A) throws IOException {
        for (int i = 0; i < A.length; i++) A[i] = scanLong();
    }

    public void scanInt(int[] A) throws IOException {
        for (int i = 0; i < A.length; i++) A[i] = scanInt();
    }

    public double scanDouble() throws IOException {
        int c = scan();
        while (isWhiteSpace(c)) c = scan();
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = scan();
        }
        double res = 0;
        while (!isWhiteSpace(c) && c != '.') {
            if (c == 'e' || c == 'E') {
                return res * Math.pow(10, scanInt());
            }
            res *= 10;
            res += c - '0';
            c = scan();
        }
        if (c == '.') {
            c = scan();
            double m = 1;
            while (!isWhiteSpace(c)) {
                if (c == 'e' || c == 'E') {
                    return res * Math.pow(10, scanInt());
                }
                m /= 10;
                res += (c - '0') * m;
                c = scan();
            }
        }
        return res * sgn;
    }

}

优点:

扫描输入比BufferReader更快 降低时间复杂度 为每个下一个输入刷新缓冲区

方法:

scanChar() -扫描单个字符 scanInt() - scan整数值 scanLong() -扫描长值 scanString() -扫描字符串值 scanouble () - scan双值 scanInt(int[] array) -扫描完整的数组(整数) scanLong(long[] array) -扫描完整的array (long)

用法:

复制java代码下面的给定代码。 初始化给定类的对象

ScanReader sc = new ScanReader(System.in); 3.导入必要的类:

进口java.io.BufferedInputStream; 进口java.io.IOException; 进口java.io.InputStream; 4. 从主方法中抛出IOException来处理Exception 5. 使用提供的方法。 6. 享受

例子:

import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
class Main{
    public static void main(String... as) throws IOException{
        ScanReader sc = new ScanReader(System.in);
        int a=sc.scanInt();
        System.out.println(a);
    }
}
class ScanReader....

那是因为扫描仪。nextInt方法不读取通过按“Enter”创建的输入中的换行符，因此调用Scanner。nextLine在读取换行符后返回。

当您使用Scanner时，您将遇到类似的行为。在Scanner.next()或任何Scanner之后的nextLine。nextFoo方法(除了nextLine本身)。

处理:

Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline int option = input.nextInt(); input.nextLine(); // Consume newline left-over String str1 = input.nextLine(); Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method. int option = 0; try { option = Integer.parseInt(input.nextLine()); } catch (NumberFormatException e) { e.printStackTrace(); } String str1 = input.nextLine();