2

我有一个文件名列表,我想在上面实现Dir::glob

我认为这很容易,因为File::fnmatch看起来是正确的工具。但是,在某些情况下,这两种方法的行为不同:

# Given a directory layout like this:
# +
# |-- file
# |-+ folder
# | |-- file

# gives ['file','folder/file']
Dir.glob('**/file') 

# gives ['folder/file']
['file','folder','folder/file'].select{|n| File.fnmatch?('**/file', n) }

添加斜线可以解决这个问题,但会引入另一个问题:

# gives ['file,'folder/file']
['file','folder','folder/file'].select{|n| File.fnmatch?('**/file','/'+n) }
# gives an empty array, but should give 'file' and 'folder' like Dir.glob does.
['file','folder','folder/file'].select{|n| File.fnmatch?('f*','/'+n) }

有人已经解决了这个问题还是我必须做一些正则表达式魔术(tm)?

4

3 回答 3

3

你需要指向一个FNM_PATHNAME标志

> ['file','folder','folder/file'].select{|n| File.fnmatch?('**/file', n, File::FNM_PATHNAME) }
 => ["file", "folder/file"] 
> ['file','folder','folder/file'].select{|n| File.fnmatch?('f*',n, File::FNM_PATHNAME) }
 => ["file", "folder"] 
于 2013-05-09T10:59:48.810 回答
2

尝试传递File::FNM_PATHNAME旗帜,

 ['file','folder','folder/file'].select{|n| File.fnmatch?('**/file', n,File::FNM_PATHNAME) }
 => ["file", "folder/file"]

这似乎给了你想要的东西..

于 2013-05-09T10:59:42.310 回答
0

不确定这是否适用于所有情况,但 **file (不带斜线)似乎与您的两个测试用例匹配:

 ['file', 'folder', 'folder/file'].select {|n| File.fnmatch?('**file', n) }
于 2013-05-09T10:56:48.267 回答