blocks|key|1167625|text|我认为你可以用FileTest.directory?+(file_name)测试每个文件，看看它是否是一个目录。有关详细信息，请参阅the+documentation+for+FileTest。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1167626|entityMap|0|LINK|mutability|MUTABLE|url|http://ruby-doc.org/core/classes/FileTest.html^0|7|V|1U|U|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

I think you can test each file to see if it is a directory with <code>FileTest.directory? (file_name)</code>. See <a href="http://ruby-doc.org/core/classes/FileTest.html" rel="nofollow noreferrer">the documentation for FileTest</a> for more info.

blocks|key|1382629|text|Jordan很接近，但是Dir.entries没有返回File.directory?期望的完整路径。试试这个：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1382630|+Dir.entries('/your_dir').select+{%7Centry%7C+File.directory?+File.join('/your_dir',entry)+and+!(entry+=='.'+%7C%7C+entry+==+'..')+}|code-block|syntax|javascript|1382631|entityMap^0|C|B|R|F|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Jordan is close, but <code>Dir.entries</code> doesn't return the full path that <code>File.directory?</code> expects. Try this:

<pre><code> Dir.entries('/your_dir').select {|entry| File.directory? File.join('/your_dir',entry) and !(entry =='.' || entry == '..') }
</code></pre>

blocks|key|1167697|text|directory+=+'Folder'
puts+Dir.entries(directory).select+{+%7Cfile%7C+File.directory?+File.join(directory,+file)}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1167698|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>directory = 'Folder'
puts Dir.entries(directory).select { |file| File.directory? File.join(directory, file)}
</code></pre>

blocks|key|1382665|text|在我看来，Pathname比普通字符串更适合于文件名。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1382666|require+"pathname"
Pathname.new(directory_name).children.select+{+%7Cc%7C+c.directory?+}|code-block|syntax|javascript|1382667|这将为您提供该目录中所有目录作为路径名对象的数组。|1382668|如果您想要字符串|1382669|Pathname.new(directory_name).children.select+{+%7Cc%7C+c.directory?+}.collect+{+%7Cp%7C+p.to_s+}|1382670|如果directory_name是绝对的，那么这些字符串也是绝对的。|1382671|entityMap^0|5|8|0|0|0|0|0|2|E|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

In my opinion <code>Pathname</code> is much better suited for filenames than plain strings.

<pre><code>require "pathname"
Pathname.new(directory_name).children.select { |c| c.directory? }
</code></pre>

This gives you an array of all directories in that directory as Pathname objects.

If you want to have strings

<pre><code>Pathname.new(directory_name).children.select { |c| c.directory? }.collect { |p| p.to_s }
</code></pre>

If <code>directory_name</code> was absolute, these strings are absolute too.

blocks|key|1382690|text|我发现这更有用，也更容易使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1382691|Dir.chdir('/destination_directory')
Dir.glob('*').select+{%7Cf%7C+File.directory?+f}|code-block|syntax|javascript|1382692|它获取当前目录中的所有文件夹，不包括.和..。|offset|length|style|CODE|1382693|要递归文件夹，只需使用**代替*即可。|1382694|还可以将Dir.glob行作为块传递给Dir.chdir：|1382695|Dir.chdir('/destination+directory')+do
++Dir.glob('*').select+{+%7Cf%7C+File.directory?+f+}
end|1382696|entityMap^0|0|0|I|1|K|2|0|B|2|F|1|0|4|8|J|9|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@$I|12|J|13|K|L]|$I|14|J|15|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@$I|17|J|18|K|L]|$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|T|$]]

I've found this more useful and easy to use:

<pre><code>Dir.chdir('/destination_directory')
Dir.glob('*').select {|f| File.directory? f}
</code></pre>

it gets all folders in the current directory, excluded <code>.</code> and <code>..</code>.

To recurse folders simply use <code>**</code> in place of <code>*</code>.

The <code>Dir.glob</code> line can also be passed to <code>Dir.chdir</code> as a block: 

<pre><code>Dir.chdir('/destination directory') do
 Dir.glob('*').select { |f| File.directory? f }
end
</code></pre>

blocks|key|1167813|text|递归查找某个目录下的所有文件夹：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1167814|Dir.glob+'certain_directory/**/*/'|code-block|syntax|javascript|1167815|非递归版本：|1167816|Dir.glob+'certain_directory/*/'|1167817|注意：Dir.[]的工作方式与Dir.glob类似。|offset|length|style|CODE|1167818|entityMap^0|0|0|0|0|3|6|F|8|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Recursively find all folders under a certain directory:

<pre><code>Dir.glob 'certain_directory/**/*/'
</code></pre>

Non-recursively version:

<pre><code>Dir.glob 'certain_directory/*/'
</code></pre>

Note: <code>Dir.[]</code> works like <code>Dir.glob</code>.

blocks|key|1382771|text|Dir.glob('/your_dir').reject+{%7Ce%7C+!File.directory?(e)}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1382772|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>Dir.glob('/your_dir').reject {|e| !File.directory?(e)}
</code></pre>

blocks|key|1167843|text|$dir_target+=+"/Users/david/Movies/Camtasia+2/AzureMobileServices.cmproj/media"

Dir.glob("#{$dir_target}/**/*").each+do+%7Cf%7C+
++if+File.directory?(f)
++++puts+"#{f}\n"
++end
end|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1167844|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>$dir_target = "/Users/david/Movies/Camtasia 2/AzureMobileServices.cmproj/media"

Dir.glob("#{$dir_target}/**/*").each do |f| 
 if File.directory?(f)
 puts "#{f}\n"
 end
end
</code></pre>

blocks|key|1167896|text|对于一般的解决方案，您可能希望使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1167897|Dir.glob(File.expand_path(path))|code-block|syntax|javascript|1167898|这将适用于像~/*/这样的路径(主目录中的所有文件夹)。|offset|length|style|CODE|1167899|entityMap^0|0|0|6|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

For a generic solution you probably want to use

<pre><code>Dir.glob(File.expand_path(path))
</code></pre>

This will work with paths like <code>~/*/</code> (all folders within your home directory).

blocks|key|1167945|text|我们可以将Borh's+answer和johannes'+answer结合起来，得到一个相当优雅的解决方案来获取文件夹中的目录名。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1167946|#+user+globbing+to+get+a+list+of+directories+for+a+path
base_dir_path+=+''
directory_paths+=+Dir.glob(File.join(base_dir_path,+'*',+''))

#+or+recursive+version:
directory_paths+=+Dir.glob(File.join(base_dir_path,+'**',+'*',+''))

#+cast+to+Pathname
directories+=+directory_paths.collect+{%7Cpath%7C+Pathname.new(path)+}

#+return+the+basename+of+the+directories
directory_names+=+directories.collect+{%7Cdir%7C+dir.basename.to_s+}|code-block|syntax|javascript|1167947|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/15473045/848668|1|https://stackoverflow.com/a/1899885/848668^0|5|D|0|J|G|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

We can combine <a href="https://stackoverflow.com/a/15473045/848668">Borh's answer</a> and <a href="https://stackoverflow.com/a/1899885/848668">johannes' answer</a> to get quite an elegant solution to getting the directory names in a folder.

<pre><code># user globbing to get a list of directories for a path
base_dir_path = ''
directory_paths = Dir.glob(File.join(base_dir_path, '*', ''))

# or recursive version:
directory_paths = Dir.glob(File.join(base_dir_path, '**', '*', ''))

# cast to Pathname
directories = directory_paths.collect {|path| Pathname.new(path) }

# return the basename of the directories
directory_names = directories.collect {|dir| dir.basename.to_s }
</code></pre>

blocks|key|1382954|text|仅文件夹('.‘和“..”不包括在内)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1382955|Dir.glob(File.join(path,+"*",+File::SEPARATOR))|offset|length|style|CODE|1382956|文件夹和文件：|1382957|Dir.glob(File.join(path,+"*"))|1382958|entityMap^0|0|0|1B|0|0|0|U|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|M|$]]

Only folders ('.' and '..' are excluded):

<code>Dir.glob(File.join(path, "*", File::SEPARATOR))</code>

Folders and files:

<code>Dir.glob(File.join(path, "*"))</code>

How do I get a list of the folders that exist in a certain directory with ruby?

<code>Dir.entries()</code> looks close but I don't know how to limit to folders only.

Getting a list of folders in a directory

如何使用ruby获取某个目录中存在的文件夹列表？Dir.entries()看起来很接近，但我不知道如何仅限于文件夹。

问获取目录中的文件夹列表
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