.val(undefined) returns jQuery obj instead of value

[jonny.fillmore@…]

In jQuery 1.4.3 the 'val' function behaves inconsistently if an undefined value is passed in explicitly. If you have a wrapper around this method that mirrors the "if param is undefined, return value" convention then it will no longer work due to 1.4.3 checking the number of parameters given, not the first param's value.

For example:

1.4.2

---

1341 val: function( value ) { 1342 if ( value === undefined ) { 1343 var elem = this[0];

item = $('input[type=text]'); item.val("foo"); value = undefined; item.val(undefined); returns "foo" item.val(value); returns "foo" item.val(); returns "foo"

1.4.3

---

1534 val: function( value ) { 1535 if ( !arguments.length ) { 1536 var elem = this[0];

item = $('input[type=text]'); item.val("foo"); value = undefined; item.val(undefined); returns [<input type="text"...>] item.val(value); returns [<input type="text"...>] item.val(); returns "foo"

[snover]

This is by design. See #4130.

[victor]

As a consequence of this issue, the proper 'val' substitution should look as follows:

(function () {
    if (!jQuery || !jQuery.fn.val)
        return;

    var original = jQuery.fn.val;
    jQuery.fn.val = function (value) {

	... do something ...

        // jQuery ticket #7360 (http://bugs.jquery.com/ticket/7360)
        return typeof(value) !== "undefined" ? original.call(this, value) : original.call(this);
    };
})();