I'm trying to find the index of a row in a table. I'm trying to use the following code, but I seem to get an index of -1.
$(document).ready(function()
{
$("tr").click(function (){
var index = $("table").index($(this));
$("span").text("That was row index #" + index);
});
});
With html that looks like this;
<table>
<tbody>
<tr><td>click</td></tr>
<tr><td>click</td></tr>
<tr><td>click</td></tr>
<tr><td>click</td></tr>
</tbody>
Thanks
Have you tried:
$("tr").index(this)
The documentation shows just passing this and that the preceding selection should be where the node is found. If you need to find it in a specific table (and there are multiple), you may need to provide some context:
// haven't tested this
$("tr", $(this).closest("table")).index(this)
this isnt necessary to be passed in the index function as it gave me -1 every time, using without this parameter works ok
Try:
var index = $("table tr").index(this);
The documentation for index() says:
Searches every matched element for the object and returns the index of the element, if found, starting with zero. If a jQuery object is passed, only the first element is checked.
You need to call the index() on a collection of <tr> elements, not the parent <table>.
$() around this, since I'm not sure if jQuery would return the same object instance on calls to $ with the same node instance.this as the parameter just remove it and use .index() it will work correctly @WinnemuccaBased on Robs answer to find index in specific table, this worked for me.
var index = $('tr', $(this).closest("table")).index(this);
I've just found an interesting trick, which basically consists on counting the previous siblings:
var tr = $(some_selector);
var rowIndex = tr.prevAll().length;
This way you will get 0 if this is the first tr, 3 if this is the 4th tr, etc.
Just for the sake of it, another option using index(), which saves you from having to know how to select the containing table:
var rowIndex = tr.parent().children().index(tr);
Get the td object and its parent index is tr index
$(this).parent().index()
