In JavaScript, everything is 'truthy' or 'falsy', and for numbers 0 means false, everything else true. So you could write:
if ($(selector).length)
You don't need that >0 part.
The approach you suggest is not guaranteed to give you the result you're looking for - what if you had a tbody for example:
<table id="myTable">
<tbody>
<tr>...</tr>
<tr>...</tr>
</tbody>
</table>
You would end up with the following:
<table id="myTable">
<tbody>
<tr>...</tr>
<tr>...</tr>
</tbody>
<tr>...</tr>
</table>
I would therefore recommend this approach instead:
$('#myTable tr:last').after('<tr>...</tr><tr>...</tr>');
You can include anything within the after() method as long as it's valid HTML, including multiple rows as per the example above.
Update: Revisiting this answer following recent activity with this question. eyelidlessness makes a good comment that there will always be a tbody in the DOM; this is true, but only if there is at least one row. If you have no rows, there will be no tbody unless you have specified one yourself.
DaRKoN_ suggests appending to the tbody rather than adding content after the last tr. This gets around the issue of having no rows, but still isn't bulletproof as you could theoretically have multiple tbody elements and the row would get added to each of them.
Weighing everything up, I'm not sure there is a single one-line solution that accounts for every single possible scenario. You will need to make sure the jQuery code tallies with your markup.
I think the safest solution is probably to ensure your table always includes at least one tbody in your markup, even if it has no rows. On this basis, you can use the following which will work however many rows you have (and also account for multiple tbody elements):
$('#myTable > tbody:last-child').append('<tr>...</tr><tr>...</tr>');
Best Answer
You can use
$(selector).position()to get the X/Y coords of the bar. It returns an object with "top" and "left" keys containing the element's full page offset.Here's a fiddle with a working example: http://jsfiddle.net/xJSMu/
PS: You got your x and y offsets the wrong way round in your fiddle - by names anyway.
PPS: I've also added some selector caching (Rule of thumb: if you use it more than once, cache it instead of calling the jQuery constructor again).