javascript: get un-destructed parameter in function

Question

I need to get the un-destructed parameters inside a function. What would be the best way?

const foo = ({url, options, header, body, auth = 1}) => {
    //...do things with the parameters...
    let params = {url, options, header, body, auth}; // Is there an easy way?
    bar(params);
}

const foo = params => { }; don't destructure if you need undestructered??? — baao, Jun 26 '18 at 12:00
But then I lost the convenience of parameter destructing. I need both :) — Xun Yang, Jun 26 '18 at 12:01
Possible duplicate of [ES6 destructuring function parameter - naming root object](https://stackoverflow.com/questions/29051011/es6-destructuring-function-parameter-naming-root-object) — Sebastian Simon, Jun 26 '18 at 12:05

Danziger · Accepted Answer · 2018-07-04T10:19:52.913

You could have a single parameter in foo and do the destructuring inside it. Then, you would do some stuff with url, options, header, body and auth and finally call bar like bar({ ...args, auth }), spreading args and adding auth as well:

const bar = (baz) => {
  console.log(baz);
};

const foo = (args) => {
  const { url, options, header, body, auth = 1 } = args;

  // Do stuff with url, options, header, body and auth...

  bar({ ...args, auth });
};

foo({ url: 'www.google.com', options: {}, header: {}, body: 'Hi there!' });
foo({ url: 'www.google.com', options: {}, header: {}, body: 'Hi there!', auth: false });

.as-console-wrapper {
  max-height: 100vh !important;
}

javascript: get un-destructed parameter in function

1 Answers1