Fileoutputstream 没有创建另一个文件[重复]答案

【问题标题】：Fileoutputstream is not creating another file [duplicate]Fileoutputstream 没有创建另一个文件[重复]
【发布时间】：2020-05-16 22:11:10
【问题描述】：

我正在尝试让我的程序创建另一个可由用户命名的文件并打印出用户想要打印的随机数。不幸的是，当我运行我的程序时，它会一直运行，直到我到达程序状态的最后一个“你想使用什么文件名”

"找不到文件 java.io.FileNotFoundException: (句柄是无效）”

我想知道如何让程序实际创建文件并让用户为文件选择自己的名称。

import java.io.*;
import java.util.*;

public class chooseRandNum{
    public static void main(String[] args){
        Random rand = new Random();
        Scanner key = new Scanner(System.in);
        System.out.println("How many random numbers do you want? ");
        int totalRand = key.nextInt();
        System.out.println("What is the smallest random number? ");
        int smallRand = key.nextInt();
        System.out.println("What is the largest random number? ");
        int largeRand = key.nextInt();
        System.out.println("What filename do you want to use? ");
        String fname = key.nextLine();

        File outputFile = new File(fname);
        PrintStream outputStream = null;

        try{
            outputStream = new PrintStream(outputFile);
        }

        catch (Exception e){
            System.out.println("File not found " + e);
            System.exit(1);
        }

        int n = rand.nextInt(largeRand - smallRand + 1);
        for(int i = 0; i <= 5; i++){
            for(int j = 0; j <= totalRand; j++){
                outputStream.print(n + ",");
            }
            outputStream.println();
        }   
    }
}

【问题讨论】：

您可能想在尝试为其打开输出流之前调用outputFile.createNewFile()。
@Powerlord 其实没必要。

标签： java output java.util.scanner printwriter

【解决方案1】：

String fname = key.nextLine(); 应该是String fname = key.next();

代码还有其他问题，但这解决了问题中的问题。

来自 JavaDoc：

nextLine()

/**
 * Advances this scanner past the current line and returns the input
 * that was skipped.
 *
 * This method returns the rest of the current line, excluding any line
 * separator at the end. The position is set to the beginning of the next
 * line.
 *
 * <p>Since this method continues to search through the input looking
 * for a line separator, it may buffer all of the input searching for
 * the line to skip if no line separators are present.
 *
 * @return the line that was skipped
 * @throws NoSuchElementException if no line was found
 * @throws IllegalStateException if this scanner is closed
 */

next()

/**
 * Finds and returns the next complete token from this scanner.
 * A complete token is preceded and followed by input that matches
 * the delimiter pattern. This method may block while waiting for input
 * to scan, even if a previous invocation of {@link #hasNext} returned
 * {@code true}.
 *
 * @return the next token
 * @throws NoSuchElementException if no more tokens are available
 * @throws IllegalStateException if this scanner is closed
 * @see java.util.Iterator
 */

【讨论】：

还有什么问题？
@JanezKuhar 这超出了问题的范围。运行它并查看输出。例如。你认为i <= 5 的嵌套循环在这里做什么？
哦，我错过了！

【解决方案2】：

就在这些陈述之后：

System.out.println("What is the largest random number? ");
int largeRand = key.nextInt();

放这个

key.nextLine(); // clear lingering new line

然后继续：


System.out.println("What filename do you want to use? ");
String fname = key.nextLine();

并在此网站上查看以下内容：Scanner usage

【讨论】：