/存储/仿真/0/AndroidAPP/addon_data/plugin.audio.mp3streams/artist_icons/David Bowie.jpg:open failed:enoint（没有这样的文件或目录）

我想在android中提取zip文件。对于其他Zip文件，它工作正常，但我有一个Zip文件，无法解压缩到SD卡的任何路径。 它给了我这个错误

  /storage/emulated/0/AndroidTVAPP/addon_data/plugin.audio.mp3streams/artist_icons/David Bowie.jpg: open failed: ENOENT (No such file or directory)

解压缩文件JAVA类

public class UnzipUtility  
{ 
/**
 * Size of the buffer to read/write data
 */
private static final int BUFFER_SIZE = 4096;
/**
 * Extracts a zip file specified by the zipFilePath to a directory specified by
 * destDirectory (will be created if does not exists)
 * @param zipFilePath
 * @param destDirectory
 * @throws IOException
 */
public void unzip(String zipFilePath, String destDirectory) throws IOException 
{
    File destDir = new File(destDirectory);
    if (!destDir.exists()) 
    {
        destDir.mkdir();
    }
    ZipInputStream zipIn = new ZipInputStream(new FileInputStream(zipFilePath));
    ZipEntry entry = zipIn.getNextEntry();
    // iterates over entries in the zip file
    while (entry != null) 
    {



        String filePath = destDirectory + File.separator + entry.getName();

        String filePath2 = File.separator + entry.getName();// filePath;

        if (null != filePath2 && filePath2.length() > 0 )
        {
            int endIndex = filePath2.lastIndexOf("/");
            if (endIndex != -1)  
            {
                filePath2 = filePath2.substring(0, endIndex); // not forgot to put check if(endIndex != -1)

                 Log.e("File path=",""+filePath2);

                 File root = android.os.Environment.getExternalStorageDirectory();
                 File dir = new File(root.getAbsolutePath() + "/AndroidTVAPP"+File.separator+filePath2);
                  if(dir.exists() == false){
                      dir.mkdirs();  
                  }



            }
        }  


        if (!entry.isDirectory()) {
            // if the entry is a file, extracts it
            extractFile(zipIn, filePath);
        } else {
            // if the entry is a directory, make the directory
            File dir = new File(filePath);
            dir.mkdir();
        }
        zipIn.closeEntry();
        entry = zipIn.getNextEntry();
    }
    zipIn.close();
}
/**
 * Extracts a zip entry (file entry)
 * @param zipIn
 * @param filePath
 * @throws IOException
 */
private void extractFile(ZipInputStream zipIn, String filePath) throws IOException 
{
    BufferedOutputStream bos = new BufferedOutputStream(new FileOutputStream(filePath));
    byte[] bytesIn = new byte[BUFFER_SIZE];
    int read = 0;
    while ((read = zipIn.read(bytesIn)) != -1) {
        bos.write(bytesIn, 0, read);
    }
    bos.close();
}

---

}

您需要使用转义字符来指定文件名中有空格

试试这个

String path = "/storage/emulated/0/AndroidTVAPP/addon_data/plugin.audio.mp3streams/artist_icons/";
String fullPath= path + "David Bowie.jpg";
fullPath = fullPath.replace(" ", "\\ ");

或者您可以使用

Uri.parse（path）

---

必须是文件名中的空格。你试过编码吗？@KNeerajLal可能有空间，因为zip文件大约有45MB，并且有很多文件，我是否应该在这里上传解压类，这样你就可以看到它，然后culd会帮助我。然后你需要。我已经修复了代码，同时我放了一些创建目录的行。上面的代码已经更新，并且工作正常。谢谢你告诉我在移动文件之前应该创建目录

public void unzip(String zipFilePath, String destDirectory) throws IOException {
    ...
    ZipInputStream zipIn = new ZipInputStream(new FileInputStream(Uri.parse(zipFilePath)));
    ...
}