Java如何获取没有全名的文件 [英] Java how to get file without full name

问题描述

我需要从应用程序的另一部分中的应用程序下载的目录中获取文件，但是此文件的名称是动态创建的.我的意思是，它始终是Query_，然后是一些数字，fe. Query_21212121212，Query_22412221.我需要获取绝对文件路径，例如C:/dir/Query_ *.该怎么做?

I need to get the file from directory which is downloaded from application in another part of app, but name of this file is creating dynamically. I mean that it is always Query_ and then some digits, fe. Query_21212121212, Query_22412221. I need to get the absolute file path, sth like C:/dir/Query_*. How to do that?

推荐答案

如果知道文件所在的路径，则可以创建DirectoryStream并获取其中的所有文件.

If you know the Path where the files reside, you can create a DirectoryStream and get all the files in it.

请参见 https://docs.oracle.com/javase/tutorial/essential/io/dirs.html#glob

示例(通过上面的链接):

Example (from link above):

Path dir = ...;
try (DirectoryStream<Path> stream =
     Files.newDirectoryStream(dir, "*.{java,class,jar}")) { //<--- change the glob to 
                                                            //fit your name pattern.
    for (Path entry: stream) {
        System.out.println(entry.getFileName());
    }
} catch (IOException x) {
    // IOException can never be thrown by the iteration.
    // In this snippet, it can // only be thrown by newDirectoryStream.
    System.err.println(x);
}

关于glob: https://docs.oracle. com/javase/tutorial/essential/io/fileOps.html#glob

因此，在您的情况下，可能类似于"Query_*".您使用特殊扩展名还是根本不使用扩展名?如果是.txt，则全局名称应类似于"Query_[0-9]+.txt".

So in your case it would probably be something like "Query_*". Do you use a special extension or no extension at all? If it were .txt then the glob should be like "Query_[0-9]+.txt".

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