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>');
Just use this:
colspan="100%"
It works on Firefox 3.6, IE 7 and Opera 11! (and I guess on others, I couldn't try)
Warning: as mentioned in the comments below this is actually the same as colspan="100"
. Hence, this solution will break for tables with css table-layout: fixed
, or more than 100 columns.
Best Answer
There are some possible solutions, but the best one would depend on your requirements and restrictions. If it's not a huge application, and you don't expect to have too many items in your unfiltered array, the best solution would be probably to simply use a
ng-show
with the same filter:But keep in mind that your filter will run through all items of the array twice, on every digest cycle. And if performance might be a problem, then you probably want your controller to digest this value for you and just bind it to your scope:
And in your HTML: