jquery - add to array

Question

Why this not working?

var inputs = new Array();
$("input").each(function(){
    input = $(this).val();
})

console.log(input);

how to correctly use arrays in jQuery? Like as PHP?

Arrays are part of native JavaScript, there's no special jQuery syntax for them. However, your code shows a total lack of understanding of basic programming concepts, so I'd suggest you go back and cover those. — Anthony Grist
– Anthony Grist, Commented Aug 7, 2012 at 11:11
@AnthonyGrist he is asking a question because he doesn't know to do it. — Ram
– Ram, Commented Aug 7, 2012 at 11:13
@Raminson He's asking a question that he shouldn't need to ask if he'd covered the basic concepts of programming in JavaScript. There are plenty of well-written guides covering them to be found in less than a minute on Google. There's no research effort on his part (criteria for downvoting) and the question isn't going to be useful to anybody else (criteria for closing as too localized). — Anthony Grist
– Anthony Grist, Commented Aug 7, 2012 at 11:34

Community · Accepted Answer · 2020-06-20 09:12:55Z

2

I assume what you are trying to do is get an array of the values of all the <input> elements on your page. What you'll need to do is iterate over all the elements using the .each() function and append each value to your inputs array.

Try this -

 var inputs = new Array();
  $("input").each(function(){
      inputs.push($(this).val());
  })

  console.log(inputs);

You need to use the push() function to add an element to an array.

fiddle demo

References -

Array push method on MDN

As a final note, here is a shorthand way to define a new array -

var inputs = [];

That line is functionally identical to -

var inputs = new Array();

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Aug 7, 2012 at 11:11

Lix

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1 Comment

Anthony Grist Over a year ago

Technically input = $(this).val(); will define the variable on the global scope.

Manse · Accepted Answer · 2012-08-07 11:11:32Z

2

Use Array.push

var inputs = new Array();
$("input").each(function(){
   inputs.push($(this).val());
})

Also note the variable differences .. input != inputs

answered Aug 7, 2012 at 11:11

Manse

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1 Comment

Ram Over a year ago

But I think this is more constructive. +1

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