How to create variable which contains function with input that are variable

Question

I want to create variable which contains a function with variable input.

I tried this but no success


var setup = function( isAjaxSuccess ) {
    return function( ajaxSuccess = isAjaxSuccess ){

       if( ajaxSuccess === true ){
           console.log("success");
       } else {
           console.log("error");
       }        

    };

};

Here we call setup( true ) or setup( false ) according to which function changes. Here console.log is just for show. Here setup( true ) should return a function basically.

What is the error you are getting? There's a lot of things which can go wrong. The variable syntax seems to be correct — sam
– sam, Commented Jun 27, 2019 at 3:34

Jack Bashford · Accepted Answer · 2019-06-27 03:38:12Z

2

Just return a function that uses ajaxSuccess:

var setup = function(ajaxSuccess) {
  return function() {
    if (ajaxSuccess === true) {
      console.log('success');
    } else {
      console.log('fail');
     }
  }
};

setup(true)();
setup(false)();

You could also make it more concise like so:

const setup = ajaxSuccess => () => ajaxSuccess ? console.log("success") : console.log("fail");
setup(true)();
setup(false)();

answered Jun 27, 2019 at 3:38

Jack Bashford

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Comments

Obsidian Age · Accepted Answer · 2019-06-27 03:35:39Z

0

You don't need to run the return statement; just check the input inside the conditional:

var setup = function(ajaxSuccess) {
  if (ajaxSuccess === true) {
    console.log('success');
  } else {
    console.log('fail');
  }

};

setup(true);
setup(false);

answered Jun 27, 2019 at 3:35

Obsidian Age

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1 Comment

rishabhjainps Over a year ago

Here setup(true) will not return a function which is what i want for my usecase

zeterain · Accepted Answer · 2019-06-27 03:37:05Z

0

Because calling setup(true) returns a function, you will need to, in turn, call that function. Try:

var setup = function(isAjaxSuccess) {
    return function() {
        if (isAjaxSuccess === true ) {
            console.log("success");
        } else {
            console.log("error");
        }
    };
};

var result = setup(true);
result(); // will log "success" to the console

answered Jun 27, 2019 at 3:37

zeterain

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