0
public static void main(String[] args) throws  IOException  {
    String filename = "C:\\audiofile.wav";
    InputStream in = null;
    try{
       in = new FileInputStream(filename);
    } 
    catch(FileNotFoundException ex){
         System.out.println("File not found");
    }
    AudioStream s = null;
    s = new AudioStream(in);
    AudioPlayer.player.start(s);
}

我已经在 netbeans 中编写了这段代码。我的音频文件的名称是 audiofile.wav。但它一直显示异常“找不到文件”。谁能帮我 ???

4

3 回答 3

2

Windows Vista 及以上驱动器中的根文件夹C受 UAC 保护。这需要您在管理模式下运行 java 可执行文件。

但是,您可以将 wav 文件转移到其他地方,UAC 不会干扰(例如Documents您当前登录用户的文件夹)或不同驱动器的根目录(例如 D:\ 和 E:)

另外,请确保audiofile.wav确实在所述位置(C:\audiofile.wav

于 2013-11-12T09:01:06.727 回答
0

我认为首先,您应该粘贴您的异常代码!
那么,我认为 java I/O 支持这两种方式:

"C:/audiofile.wav"
"C:\audiofile.wav"

导入 java.io.BufferedReader;导入 java.io.FileInputStream;导入 java.io.InputStreamReader;

公共类主要{

public static void main(String[] args) {
    // write your code here
    String fileLocation = "C:\\1.diff";
    String fileLocation1 = "C:/1.diff";
    try {
        FileInputStream f = new FileInputStream(fileLocation);

        BufferedReader reader = new BufferedReader(new InputStreamReader(f));

        String line = reader.readLine();
        System.out.println("11111111111111111111111111");

        while (line != null) {
            // Process line
            line = reader.readLine();
            System.out.println(line);

        }
        System.out.println("11111111111111111111111111");


    } catch (Exception ex) {

        System.out.println(ex);
    }
    try {
        FileInputStream ff = new FileInputStream(fileLocation1);
        BufferedReader reader1 = new BufferedReader(new InputStreamReader(ff));
        String line1 = reader1.readLine();
        System.out.println("2222222222222222222222222");

        while (line1 != null) {
            // Process line
            line1 = reader1.readLine();
            System.out.println(line1);

        }
        System.out.println("2222222222222222222222222");

    } catch (Exception ex) {

        System.out.println(ex);
    }

}

}

有用。我不知道你做了什么,无论如何粘贴你的错误消息!

====

```
    private static void B() {
        String filename = "C:\\test.wav";
        InputStream in = null;
        try {
            in = new FileInputStream(filename);
        } catch (FileNotFoundException ex) {
            System.out.println("File not found");
        }
        try {
            AudioStream s = new AudioStream(in);
            AudioPlayer.player.start(s);
        } catch (IOException ex) {
            System.out.println(ex.getMessage());
        }

    }

```

有用!

于 2013-11-13T07:00:54.083 回答
0

尝试将您的文件放在不同的位置,看看会发生什么

ProjectRootDir
          audiofile.wav
          src 

并运行这个字符串

String filename = "audiofile.wav";
于 2013-11-13T07:07:48.923 回答