What is the current directory used by fs module functions?
It's the directory where the node interpreter was started from (aka the current working directory), not the directory of script's folder location (thank you robertklep).
Also, current working directory can be obtained by:
process.cwd()
For more information, you can check out: https://nodejs.org/api/fs.html
EDIT: Because you started app.js by node app.js
at ExpressApp1
directory, every relative path will be relative to "ExpressApp1" folder, that's why fs.createWriteStream("abc")
will create a write stream to write to a file in ExpressApp1/abc
. If you want to write to ExpressApp1/routes/abc
, you can change the code in members.js
to:
fs.createWriteStream(path.join(__dirname, "abc"));
For more:
https://nodejs.org/docs/latest/api/globals.html#globals_dirname
https://nodejs.org/docs/latest/api/path.html#path_path_join_paths
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Old Geezer
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Updated on July 09, 2022Comments
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Old Geezer almost 2 years
What is the current directory when a method in the
fs
module is invoked in a typical node.js/Express app? For example:var fs = require('fs'); var data = fs.readFile("abc.jpg", "binary");
What is the default folder used to locate abc.jpg? Is it dependent on the containing script's folder location?
Is there a method to query the current directory?
My file structure is:
ExpressApp1/ app.js routes/ members.js
In members.js I have a
fs.createWriteStream("abc")
and fileabc
was created inExpressApp1/
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robertklep over 7 yearsIt's not the directory where the script is located, it's the directory where the
node
interpreter was started from (aka the current working directory). -
Old Geezer over 7 years@ardilgulez I have expanded my question with my experience.