Node js Get folder path from a file

Question

Is there a way to get the path to a folder that holds a particular file.

fs.realpathSync('config.json', []);

returns something like

G:\node-demos\7-node-module\demo\config.json

I just need

G:\node-demos\7-node-module\demo\ 
or
G:\node-demos\7-node-module\demo\

Is there any api for this or will I need to process the string?

David Murdoch · Accepted Answer · 2019-11-14 23:22:49Z

240

use path.dirname

// onlyPath should be G:\node-demos\7-handlebars-watch\demo
var onlyPath = require('path').dirname('G:\\node-demos\\7-node-module\\demo\\config.json');

edited Nov 14, 2019 at 23:22

David Murdoch

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answered Jun 19, 2013 at 13:18

hereandnow78

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6 Comments

Константин Ван Over a year ago

Double your backslashes, or you’ll end up escaping random characters.

Kip Over a year ago

if you're starting with a relative path like in the original question, you would do- let onlyPath = path.dirname(fs.realpathSync('config.json'));

joshweir Over a year ago

I would want to always use @Kip method of using realpathSync which covers absolute and relative paths

webnoob Over a year ago

Just to add a warning on this. If your path doesn't have a file name on the end (i.e your using the same code for multiple things) it will just strip off the last directory in your path instead. It's a dumb function which works well if you're 100% sure the file path has a file in it.

doug65536 Over a year ago

@webnoob You could see it as it being dumb, or you could see it as it giving the directory that contains that directory. It's not wrong. That "dumb" behaviour is exactly what you want. Did you want it to be impossible to get the name of the directory that contains a directory?

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Константин Ван · Accepted Answer · 2019-07-20 16:33:10Z

7

require("path").dirname(……) breaks when your path does not explicitly specify its directory.

require("path").dirname("./..")
// "."

You may consider using require("path").join(……, "../") instead. It preserves the trailing separator as well.

require("path").join("whatever/absolute/or/relative", "../")
// "whatever/absolute/or/" (POSIX)
// "whatever\\absolute\\or\\" (Windows)

require("path").join(".", "../")
// "../" (POSIX)
// "..\\" (Windows)

require("path").join("..", "../")
// "../../" (POSIX)
// "..\\..\\" (Windows)

require("path").win32.join("C:\\", "../")
// "C:\\"

answered Jul 20, 2019 at 16:33

Константин Ван

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