find a element with certain name in a js variable filled by a jquery query

Question

I can search the dom with .find('.dbfieldname') to find all elements from the class dbname. I also can save the result as an object in a variable like that: myelems = .find(...

But I can't search in my object for certain elements using find(). I tried the following:

my.find(*[name=versionnumber])

How can I do that right ?

This is missing so much information. What is my for starters? — Jamiec
– Jamiec, Commented Aug 9, 2011 at 7:48
my is a javascript array, which holds all the elements found by the following jquery query: my = $('input') — Andreas
– Andreas, Commented Aug 9, 2011 at 8:08

rkw · Accepted Answer · 2011-08-09 08:06:42Z

4

Your logic is fine, my is already a jQuery object, so my.find() would normally work.

The problem is that .find() searches for childrens. What you want is my.filter('[name="' +versionnumber+ '"]');

answered Aug 9, 2011 at 8:06

rkw

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2 Comments

Andreas Over a year ago

The following worked fine: my.filter('[name="' +versionnumber+ '"]'); But I'm not sure if my really is a object - I think it is an ordinary js array, because I can reach the elements by my[index]. Thank you very much.

rkw Over a year ago

Actually, I'm not really sure how it's stored. I'm going to ask SO if it's not already asked. All I know is you can retrieve the HTML elements using my[index]; but if you wanted to retrieve them as jQuery objects, you can use my.eq(index).

J0HN · Accepted Answer · 2011-08-09 07:48:24Z

0

jQuery(my).find("*[name=versionnumber]")

Btw, better use jQuery.find() or $.find() rather than just .find(), it makes code at least more readable.

answered Aug 9, 2011 at 7:48

J0HN

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2 Comments

Andreas Over a year ago

thanks, but this doesn't work as well. I tried: jQuery(my).find("") jQuery(my).find("*[name=database]") $(my).find("*[name=database]") don't know why - I can get elements like that: *my[0].name but then I would have to loop through all elements two times.

J0HN Over a year ago

Well, what is "my"? It should be a jQuery object or DOM node. If it is some hand-made JS-object you cant use jQuery on it

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