如何在不关闭Java GUI的情况下释放文件

Kevin 发表于 Dev

凯文

我创建了一个非常简单的Java GUI，以在Windows平台上浏览/加载zip文件，开始解压缩，然后进行一些文件检查。

一切正常，除了必须关闭GUI窗口以删除已在GUI中打开的zip文件之外，在解压缩方法的finally块中，我尝试添加以下内容：

public static String unZip(String path)
{
    int count = -1;
    String savepath = "";
    File file = null;
    InputStream is = null;
    FileOutputStream fos = null;
    BufferedOutputStream bos = null;
    savepath = path.substring(0, path.lastIndexOf("\\")) + File.separator; //File saving directory
    new File(savepath).mkdir(); //create the saving directory
    ZipFile zipFile = null;
    String topLevelDirName="";
    try
    {
        zipFile = new ZipFile(path,Charset.forName("gbk")); //Encoding
        Enumeration<?> entries = zipFile.entries();
        int levelCount=0;
        while(entries.hasMoreElements())
        {
            byte buf[] = new byte[buffer];
            ZipEntry entry = (ZipEntry)entries.nextElement();
            String filename = entry.getName();
            boolean ismkdir = false;
            if(filename.lastIndexOf("/") != -1){ //To check if there is a directory
                ismkdir = true;
            }
            filename = savepath + filename;
            if(entry.isDirectory()){ //If it is a directory
                levelCount++;
                file = new File(filename);
                file.mkdirs();
                if(levelCount==1)
                    topLevelDirName = filename;
                continue;
            }

            file = new File(filename);
            if(!file.exists()){
                if(ismkdir){
                    new File(filename.substring(0, filename.lastIndexOf("/"))).mkdirs();
                }
            }


            file.createNewFile(); //Create the file
            is = zipFile.getInputStream(entry);
            fos = new FileOutputStream(file);
            bos = new BufferedOutputStream(fos, buffer);
            while((count = is.read(buf)) > -1)
            {
                bos.write(buf, 0, count);
            }
            bos.flush();
            bos.close();
            fos.close();
            is.close();
        }
        zipFile.close();
    }catch(IOException ioe){
        ioe.printStackTrace();
    }finally{
        try{
            if(bos != null){
                bos.close();
            }
            if(fos != null) {
                fos.close();
            }
            if(is != null){
                is.close();
            }
            if(zipFile != null){
                zipFile.close();
            }

        }catch(Exception e) {
            e.printStackTrace();
        }
        return topLevelDirName;
    }
}

但是，除非明确关闭GUI，否则我仍然无法删除该zip。

想知道是否与Windows文件句柄有关吗？

疯狂程序员

Java 8引入了try-with-resources语句，以使这种情况更加简单和整洁。

您遇到的问题之一是，如果关闭已打开的许多资源的任何尝试失败，那么其他任何一个都不会关闭

public static String unZip(String path) throws IOException {
    int count = -1;

    File sourceFile = new File(path);
    String name = sourceFile.getName();
    name = name.substring(0, name.lastIndexOf(".zip"));
    File sourcePath = new File(sourceFile.getParent(), name);

    System.out.println("SavePath = " + sourcePath);
    if (!sourcePath.exists() && !sourcePath.mkdirs()) {
        throw new IOException("Could not create directory " + sourcePath);
    }
    String topLevelDirName = "";
    try (ZipFile zipFile = new ZipFile(sourceFile)) {
        Enumeration<?> entries = zipFile.entries();
        int levelCount = 0;
        byte buf[] = new byte[1024];
        while (entries.hasMoreElements()) {
            ZipEntry entry = (ZipEntry) entries.nextElement();
            String filename = entry.getName();
            File file = new File(sourcePath, filename);
            if (entry.isDirectory()) { //If it is a directory
                levelCount++;
                System.out.println("Make directory " + file);
                if (!file.exists() && !file.mkdirs()) {
                    throw new IOException("Could not create directory " + filename);
                }
            } else {
                System.out.println("Extract to " + file);
                try (InputStream is = zipFile.getInputStream(entry);
                                BufferedOutputStream bos = new BufferedOutputStream(new FileOutputStream(file))) {
                    while ((count = is.read(buf)) > -1) {
                        bos.write(buf, 0, count);
                    }
                }
            }
        }
    }
    return topLevelDirName;
}

我已经稍微更新了代码，以使其变得更简洁，更简单，并利用可用的API。

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