Question

118

Why is .every() not a function?

rated 0 times [ 124] [ 6] / answers: 1 / hits: 16578 / 7 Years ago, sat, april 29, 2017, 12:00:00

I've gathered an Array (I think) of required form elements, and have added 'blur' listener.

    var formInputs = $(':input').filter('[required]');

  formInputs.each(function(i) {

    $(this).on('blur', function() { // Each time we leave a 'required' field, check to see if we can activate the 'submit' button.

      submitEnabler(formInputs);

    });

  });

So, once someone has left one of these fields, I want to run through this array using .every() and check if the fields are valid - that is if they have a 'success' class that I have defined.

function isValid(input) {

  return input.hasClass('is_glowing_success');

}



function submitEnabler(inputs) {



  console.log(inputs.every(isValid));

}

I keep getting back:

Uncaught TypeError: inputs.every is not a function

    at submitEnabler

Now, I could do something like this...

for (var i = 0; i < inputs.length; i++) {

    if ($(inputs[i]).hasClass('is_glowing_success')) {

      console.log('yes');

    } else {

      console.log('no');

    }

  }

But, why can't I just use: Array.Prototype.every() ?

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2 Years ago

answered 7 Years ago jarod · Accepted Answer

jQuery does not have a .every() method. .every is defined at Array.prototype.

You can use .toArray() to convert jQuery object to an Array, within .every() callback function pass current DOM element to jQuery() to get jQuery object representation of element where .hasClass() can be chained.

function submitEnabler(inputs) {

  console.log(inputs.toArray().every(isValid));

}



function isValid(input) {

  return $(input).hasClass('is_glowing_success');

}