CODEWITHSUNDEEP
javascriptphpjquerystringforms
kexxcream
kexxcream

Reputation: 5933

Remove comma from jQuery array

Problem:

I have an array in JS that looks like this:

["intelligence", "skin", "weight", "volume"]

This is stored in a var called "newListASource". I apply this var to a form and then submit it using POST.

// Form destination
var formUrl     = 'explorer.php',

// Generate the form
$form = $("<form action='"+ formUrl +"' method='post'></form>")
        .append("<input type='hidden' name='Perspective' value='" + newListASource + "'>")
        .append("<input type='hidden' name='form_submitted' value='true'>")
        .appendTo('body');

// Submit the form
$form.submit();

When I pick it up through PHP and print it out, it gives me a string with commas. For instance:

skin,weight,intelligence,volume

Desired output I would like is (with trimmed spaces in the beginning and end):

skin weight intelligence volume

Upvotes: 3

Views: 6876

Answers (7)

nl-x
nl-x

Reputation: 11832

What you can also do is just send the array as an array!

var formUrl     = 'explorer.php';
var $form = $("<form action='"+ formUrl +"' method='post'></form>")
    .append("<input type='hidden' name='form_submitted' value='true'>")
    .appendTo('body');
while(newListASource.length)
    $form.append($("<input type='hidden' name='Perspective[]' value=''>").val(newListASource.shift()));
$form.submit();

Php will then actually have an array inside $_POST['Perspective']. (Beware the input name is now Perspective[ ] to indicate it will be an array.)

Upvotes: 0

MD Sayem Ahmed
MD Sayem Ahmed

Reputation: 29166

newListASource.join(' ') will give you the desired result - 

// Generate the form
$form = $("<form action='"+ formUrl +"' method='post'></form>")
    .append("<input type='hidden' name='Perspective' value='" 
        + newListASource.join(' ') + "'>")
    .append("<input type='hidden' name='form_submitted' value='true'>")
    .appendTo('body');

// Submit the form
$form.submit();

See the doc.

Currently you are getting comma separated values because newListASource.toString() is called whenever you are concatenating your array with a string, and this is the default behavior of toString method for arrays. When you call join, however, it converts your array to a string, separated by the delimiter that you pass to it as argument. So you get values separated by spaces in this case.

Upvotes: 5

Jai
Jai

Reputation: 74738

You can use .join() and $.trim():

// Form destination
var formUrl     = 'explorer.php',
    newListASource = newListASource.join(' ');

then in the form you can change to this:

$.trim(newListASource)

Upvotes: 2

Suvash sarker
Suvash sarker

Reputation: 3170

Simply like that:

arr.join("")

Upvotes: 0

Pir Abdul
Pir Abdul

Reputation: 2334

 var arr = ["intelligence", "skin", "weight", "volume"];


 var result = ary.join(',');

Upvotes: 0

Tushar Gupta - curioustushar
Tushar Gupta - curioustushar

Reputation: 57095

Use .join()

newListASource.join(' ')

Upvotes: 3

Arun P Johny
Arun P Johny

Reputation: 388316

Since you are using an array use Array.join() with the separator as ' '

"<input type='hidden' name='Perspective' value='" + newListASource.join(' ') + "'>"

Upvotes: 5

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