java - 如何确定在 Java 中创建文件时允许的最大路径长度

Question

如何确定在 Java 中创建文件时允许的最大路径长度。 我使用的是 Java 7，所以如果有帮助可以使用 Java NIO2，但是我如何确定文件系统允许的文件最大长度并确保我不会尝试创建具有这种无效路径的文件。

我想在尝试创建它之前缩短路径(子字符串)而不是处理之后的问题，特别是因为异常/错误消息可能因文件系统而异。

我不想对每个平台的长度进行硬编码，因为 Windows 应用程序可以访问 OSX 文件系统等等，这无论如何都行不通。

只有在尝试创建文件时才会失败 即

 try
        {

            Path p = Paths.get("C:/User/Mesh/kkkkkkkkkkkkkkkkkkkkkkkkkkrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.txt");
            Files.createFile(p);
        }
        catch(Exception e)
        {
            e.printStackTrace();
        }

给予

java.nio.file.FileSystemException: C:\User\Mesh\kkkkkkkkkkkkkkkkkkkkkkkkkkrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.txt: The filename, directory name, or volume label syntax is incorrect.

    at sun.nio.fs.WindowsException.translateToIOException(WindowsException.java:86)
    at sun.nio.fs.WindowsException.rethrowAsIOException(WindowsException.java:97)
    at sun.nio.fs.WindowsException.rethrowAsIOException(WindowsException.java:102)
    at sun.nio.fs.WindowsFileSystemProvider.newByteChannel(WindowsFileSystemProvider.java:229)
    at java.nio.file.Files.newByteChannel(Files.java:315)
    at java.nio.file.Files.createFile(Files.java:586)

与 Java 6 方法比较

try
        {
            //File file = new File("C:/User/Mesh/kkkkkkkkkkkkkkkkkkkkkkkkkkrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.txt");
            Path p = Paths.get("C:/User/Mesh/kkkkkkkkkkkkkkkkkkkkkkkkkkrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.txt");
            p.toFile().createNewFile();
            //boolean result=file.isFile();
            //System.out.println("Result is:"+result);
        }
        catch(Exception e)
        {
            e.printStackTrace();
        }

给出完全不同的错误

java.io.IOException: The filename, directory name, or volume label syntax is incorrect
    at java.io.WinNTFileSystem.createFileExclusively(Native Method)
    at java.io.File.createNewFile(File.java:947)

但真的无关紧要，因为我想在创建文件之前做一些事情

编辑: 看起来 255 个字符是一个不错的默认值 http://en.wikipedia.org/wiki/Filename无论如何

Answer 1

在 java.nio.file.FileSystem 中有一个方法 getPath 可以用来创建一个有效的路径名，否则它会抛出一个 InvalidPathException。也许这可以帮助你。