CODEWITHSUNDEEP
javascriptjqueryjoin
Edward Tanguay
Edward Tanguay

Reputation: 193312

Why doesn't .join() work with function arguments?

Why does this work (returns "one, two, three"):

var words = ['one', 'two', 'three'];
$("#main").append('<p>' + words.join(", ") + '</p>');

and this work (returns "the list: 111"):

var displayIt = function() {
    return 'the list: ' + arguments[0];
}   
$("#main").append('<p>' + displayIt('111', '222', '333') + '</p>');

but not this (returns blank):

var displayIt = function() {
    return 'the list: ' + arguments.join(",");
}   
$("#main").append('<p>' + displayIt('111', '222', '333') + '</p>');

What do I have to do to my "arguments" variable to be to use .join() on it?

Upvotes: 74

Views: 44913

Answers (9)

aggregate1166877
aggregate1166877

Reputation: 3150

You can also use Array.from and then join() with whatever token you'd like:

function joinArgs() {
  return Array.from(arguments).join(' ');
}

joinArgs('one', 'two', 'three'); // returns 'one two three'

Upvotes: 1

Ryan White
Ryan White

Reputation: 2446

At the moment you can't join array arguments, because they aren't an array, shown here

so you have to either first turn them into an array like this,

function f() {
  var args = Array.prototype.slice.call(arguments, f.length);
  return 'the list: ' + args.join(',');
}

or like this, a bit shorter

function displayIt() {
  return 'the list: ' + [].join.call(arguments, ',');
}

if you are using something like babel or a compatible browser to use es6 features, you can also do this using rest arguments.

function displayIt(...args) {
  return 'the list: ' + args.join(',');
}

displayIt('111', '222', '333');

which would let you do even cooler stuff like

function displayIt(start, glue, ...args) {
  return start + args.join(glue);
}

displayIt('the start: ', '111', '222', '333', ',');

Upvotes: 2

SpYk3HH
SpYk3HH

Reputation: 22570

You could use this jQuery .joinObj Extension/Plugin I made.

As you'll see in that fiddle, you can use it as follows:

$.joinObj(args, ",");

or

$.(args).joinObj(",");

Plugin Code:

(function(c){c.joinObj||(c.extend({joinObj:function(a,d){var b="";if("string"===typeof d)for(x in a)switch(typeof a[x]){case "function":break;case "object":var e=c.joinObj(a[x],d);e!=__proto__&&(b+=""!=b?d+e:e);break;default:"selector"!=x&&"context"!=x&&"length"!=x&&"jquery"!=x&&(b+=""!=b?d+a[x]:a[x])}return b}}),c.fn.extend({joinObj:function(a){return"object"===typeof this&&"string"===typeof a?c.joinObj(this,a):c(this)}}))})(jQuery);

Upvotes: 3

Bernardo Amorim
Bernardo Amorim

Reputation: 343

I don't know if there's a simple way to convert arguments into an array, but you can try this:

var toreturn = "the list:";
for(i = 0; i < arguments.length; i++)
{
   if(i != 0) { toreturn += ", "; }
   toreturn += arguments[i];
}

Upvotes: 1

Christian C. Salvad&#243;
Christian C. Salvad&#243;

Reputation: 827466

If you are not interested on other Array.prototype methods, and you want simply to use join, you can invoke it directly, without converting it to an array:

var displayIt = function() {
    return 'the list: ' + Array.prototype.join.call(arguments, ',');
};

Also you might find useful to know that the comma is the default separator, if you don't define a separator, by spec the comma will be used.

Upvotes: 23

easeout
easeout

Reputation: 8734

arguments is not a jQuery object, just a regular JavaScript object. Extend it before you try to call .join(). I think you would write:

return 'the list:' + $(arguments)[0];

(I'm not too familiar with jQuery, only Prototype, so I hope this is not completely bogus.)

Edit: It's wrong! But in his response, Doug Neiner describes what I'm trying to accomplish.

Upvotes: 0

Peter McG
Peter McG

Reputation: 19035

You can use typeof to see what's happening here:

>>> typeof(['one', 'two', 'three'])
"object"
>>> typeof(['one', 'two', 'three'].join)
"function"
>>> typeof(arguments)
"object"
>>> typeof(arguments.join)
"undefined"

Here you can see that typeof returns "object" in both cases but only one of the objects has a join function defined.

Upvotes: 1

John Feminella
John Feminella

Reputation: 311536

It doesn't work because the arguments object is not an array, although it looks like it. It has no join method:

>>> var d = function() { return '[' + arguments.join(",") + ']'; }
>>> d("a", "b", "c")
TypeError: arguments.join is not a function

To convert arguments to an array, you can do:

var args = Array.prototype.slice.call(arguments);

Now join will work:

>>> var d = function() {
  var args = Array.prototype.slice.call(arguments);
  return '[' + args.join(",") + ']';
}
>>> d("a", "b", "c");
"[a,b,c]"

Alternatively, you can use jQuery's makeArray, which will try to turn "almost-arrays" like arguments into arrays:

var args = $.makeArray(arguments);

Here's what the Mozilla reference (my favorite resource for this sort of thing) has to say about it:

The arguments object is not an array. It is similar to an array, but does not have any array properties except length. For example, it does not have the pop method. ...

The arguments object is available only within a function body. Attempting to access the arguments object outside a function declaration results in an error.

Upvotes: 95

Doug Neiner
Doug Neiner

Reputation: 66191

Just use the jQuery utility function makeArray

arguments is not an Array, it is an object. But, since it so "array-like", you can call the jQuery utility function makeArray to make it work:

var displayIt = function() {
    return 'the list: ' + $.makeArray(arguments).join(",");
}   
$("#main").append('<p>' + displayIt('111', '222', '333') + '</p>');

Which will output:

<p>the list: 111,222,333</p>

Upvotes: 2

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