Thursday, June 29, 2017

Module using another function of the same module

By Hường Hana 2:00 PM , , Leave a Comment

I have something like this:

  • MyModule
    • index.js
    • myFunction1.js
    • myFunction2.js

In index.js file i'm exporting all modules function (myFunction1 and myFunction2). But, in myFunction2 I use myFunction1.

If I import the index (all the module) and call it like MyModule.myFunction1 inside myFunction2, I get an error (function does not exists).

If I import myFunction1.js and call it like myFunction1(), I can't make an Stub of it when I'm going to test it.

Any aproach to do something like this?

6 Answers

Answers 1

////////////////////// // File: ./index.js // ////////////////////// const myFunction1 = require('./myFunction1.js'); const myFunction2 = require('./myFunction2.js');  module.exports = {     myFunction1,     myFunction2 };  //////////////////////////// // File: ./myFunction1.js // //////////////////////////// module.exports = function1(x){     return x; };  //////////////////////////// // File: ./myFunction2.js // //////////////////////////// const myFunction1 = require('./myFunction1.js'); // or dropping the `.js` require('./myFunction1') module.exports = function2(x){     return myFunction1(x); };

Answers 2

index.js

let fnk1 = (a) => {   return a + ' 1 ' } let fnk2 = (a) => {   return fnk1(a) } module.exports = {  fnk1,  fnk2 }

test.js

let Mymodule = require('./index') console.log(Mymodule.fnk2('argValue'));

Answers 3

This should do the trick. It would be better if they were in the same file. 

module.exports.myFunction1 = function() { ... }  module.exports.myFunction2 = function() {    module.exports.myFunction1() }

If you need them to be in separate files, you can require function1 inside of the function2 file. The thing is, each file is its own module. From the Node.js docs:

Node.js has a simple module loading system. In Node.js, files and modules are in one-to-one correspondence (each file is treated as a separate module).

Answers 4

Diego's answer works fine, except that OP seems to want to use index.js in myFunction2.js. Use the same code as Diego's for index.js

for index.js (same as Diego's)

const myFunction1 = require('./myFunction1.js'); const myFunction2 = require('./myFunction2.js');  module.exports = {     myFunction1,     myFunction2 };

for myFunction1.js (small fix)

module.exports = function function1(x){     return x; };

for myFunction2.js (changed to include index.js):

const MyModule = require('./index.js');  module.exports = function function2(x){     return MyModule.myFunction1(x); };

and a test.js :

const myFunction2 = require('./myFunction2.js');  console.log(myFunction2(20))

Answers 5

const myFunction1 = require('./myFunction1.js');  const myFunction2 = require('./myFunction2.js');  module.exports = {      myFunction1:myFunction1,      myFunction2:function(x){        return myFunction2(x,myFunction1);      }  }  ////////////////////////////  // File: ./myFunction1.js //  ////////////////////////////  module.exports = function1(x){      return x;  };    ////////////////////////////  // File: ./myFunction2.js //  ////////////////////////////    module.exports = function2(x,myFunction1){      return myFunction1(x);  };

Answers 6

You problem is not clear because you did not post your code. Since, here is a working solution for your problem:

MyModule.js:

function myFunction1() {   console.log("- I am myFunction1") }  function myFunction2() {   console.log("- I am myFunction2")   myFunction1() }  module.exports = {   myFunction1,   myFunction2 }

index.js

const MyModule = require('./MyModule')  console.log("Call myFunction1") MyModule.myFunction1() console.log("Call myFunction1") MyModule.myFunction2()

If you execute node index.js, you get the following output:

Call myFunction1 - I am myFunction1 Call myFunction2 - I am myFunction2 - I am myFunction1
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