Proper way to push to JS array

Question

I have an array called 'inputs' that I want to push some values to.

I am wondering if this is the correct way to do so:

inputs.push(name: $('#uploaderName').name(), value: $('#uploaderName').val());

#uploaderName is the ID of a form field with the value I want pushed.

FYI, the 'inputs' array is created like this:

var inputs = data.context.find(':input').not(':button');

I have also tried this with no luck:

inputs.push({name: $('#uploaderName').attr("name"), value: $('#uploaderName').val()});

To explain a bit more, at the end of this script it calls:

data.formData = inputs.serializeArray();

and then that data is passed off to a PHP script. I need to be able to add a few values to 'inputs' before inputs.serializeArray() is called.

inputs is made up of some field names and the values, I just need to be able to add a few more name/value pairs to the already generated 'inputs' variable. — Sherwin Flight
– Sherwin Flight, Commented Mar 8, 2014 at 6:56
Are you using a library like jQuery or Zepto? .find(':input').not(':button') to me suggests that data.context and inputs will be array-like Element collections rather than actual Arrays. If so, they won't have .push() methods. But, you can get an Array from them with .get(). — Jonathan Lonowski
– Jonathan Lonowski, Commented Mar 8, 2014 at 7:02

Jonathan Lonowski · Accepted Answer · 2014-03-08 07:06:55Z

1

For jQuery collections, you can include additional Elements with .add().

inputs = inputs.add('#uploaderName');

jQuery collections aren't actually Arrays. They're just array-like, meaning they have a length and numeric properties, but they won't have Array methods like .push().

answered Mar 8, 2014 at 7:06

Jonathan Lonowski

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5 Comments

Sherwin Flight Over a year ago

Doesn't seem to work. Is there any way to tell what type of data 'inputs' is so we know how to properly add to it?

Jonathan Lonowski Over a year ago

@SherwinFlight Sorry. Just updated the snippet. .add() creates a revised collection rather than mutating the original, so you have to store the result.

Sherwin Flight Over a year ago

Should it be '#uploaderName' like you have or $('#uploaderName')?

Jonathan Lonowski Over a year ago

@SherwinFlight Both should work. .add() can accept either a selector or a jQuery object.

Sherwin Flight Over a year ago

The way you had it in your answer worked fine. I am accepting your answer. Thanks for the help :)

Suman Bogati · Accepted Answer · 2014-03-08 06:41:20Z

0

You are storing the object, so use {...} to make the object first, and store it into inputs array, like

inputs.push({name: $('#uploaderName').name(), value: $('#uploaderName').val()});
//----------^----------------------------------------------------------------^

Or

var myObj  = {name: $('#uploaderName').name(), value: $('#uploaderName').val()}; 
inputs.push(myObj);

answered Mar 8, 2014 at 6:41

Suman Bogati

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1 Comment

Sherwin Flight Over a year ago

This doesn't seem to work, but I added some info to my question

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