为什么require和fs.existSync使用不同的相对路径

Question

I have this code here:

if(fs.existsSync('./example/example.js')){
    cb(require('../example/example.js'));
}else{
    cb();
}

Why should fs.existSync be using a different directory than require ?

This would be the directory tree excluding things not needed... (I am using express btw)

\example
    example.js
\routes
    index.js <-- this is the one where I am using this code
app.js <-- this one requires index.js and calls its functions using app.get('/example',example.index);

Answer 1

The path you use for require is relative to the file in which you call require (so relative to routes/index.js ); the path you use for fs.existsSync() (and the other fs functions) is relative to the current working directory (which is the directory that was current when you started node , provided that your app doesn't execute fs.chdir to change it).

As for the reason of this difference, I can only guess, but require is a mechanism for which some 'extra' logic wrt finding other modules makes sense. It should also not be influenced by runtime changes in the application, like the aforementioned fs.chdir .

Answer 2

Since the relative path to a file is relative to process.cwd(), as mentioned in this link . You can use path.resolve to resolve the path relative to the location as below:

const path = require('path');

const desiredPath = path.resolve(__dirname, './file-location');
console.log(fs.existsSync(desiredPath)); // returns true if exists