JQuery method of calculating the correct offsetLeft of an element in a table?

Var offset = $('input', myTableRow).offset(); // access offset. Left and offset. Top properties You can use this value to position other elements at the same spot: $(myTooltipElement).

AppendTo(document. Body). Css(offset).

Roatin I have one question though, suppose that it's not just contained in one table...say its contained in about 5 or 6 layers of tables (thanks firebug! ), then how do you calculate it? – leeand00 Sep 11 '09 at 19:45 Oh wait I know...I just keep passing in elements right?

– leeand00 Sep 11 '09 at 19:46 Not sure I know what you mean. Getting to the right element boils down to using the right selector. The one in my answer will actually find all input elements inside that tr, anywhere, even nested inside a table inside another table.

You need to change it to get the exact input(s) you need. – Roatin Marth Sep 11 '09 at 20:04.

Try this: var offset = $(myobject).position(). Left; $(mytooltip). Css('left',offset + 'px'); Of course you'd have to have the tooltip's position set to absolute and top (or bottom) position defined.

– Crescent Fresh Sep 11 '09 at 21:37 Oops I messed up and I meant that it needs to be defined - I was thinking about something else IE does LOL. – fudgey Sep 11 '09 at 22:06.

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