[jQuery] Re: JQuery tablesorter problem

2010-02-09 Thread Nagaraju
Hi Aquaone I am using Version 2.0.3 On Feb 4, 11:20 pm, aquaone aqua...@gmail.com wrote: Please provide the code you're using to invoke tablesorter. It's likely a syntax error. On Thu, Feb 4, 2010 at 05:36, Nagaraju man25...@gmail.com wrote: I am facing problem with jquery.tablesorter.js

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread Genus Project
when the table is generated by server side code, are you sure you are calling the correct selector (#+serverIdPrefix+table1) ? maybe you missed some letter or something. you can use firebug to examine the generated table html to see if you are in fact calling the correct selector. If you are, it

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread MorningZ
in the runat=server version, put alert($(#+serverIdPrefix+table1).length)); right before the tablesorter line believe me, as a .NET programmer myself, the runat=server is *not* causing tablesorter (or jQuery) to break, you definitely are not jQuery-selecting the table properly On Jan 12,

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread Varun Khatri
Can I also use table sorter with datagrid control? Thanks Varun On Mon, Jan 12, 2009 at 4:26 AM, MorningZ morni...@gmail.com wrote: in the runat=server version, put alert($(#+serverIdPrefix+table1).length)); right before the tablesorter line believe me, as a .NET programmer myself,

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread Varun Khatri
Also, to get around the problem : I made a div around the table : div id=tableDiv table id=table1 cellspacing=1 class=tablesorter // table content here /table /div I used this jquery after making div: $(#tableDiv).each(function(){ $(this).find(table).each(function() {

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread Varun Khatri
alert($(#+serverIdPrefix+table1).length)); This returns 1 I dint get it Plz help Thanks Varun On Mon, Jan 12, 2009 at 4:26 AM, MorningZ morni...@gmail.com wrote: in the runat=server version, put alert($(#+serverIdPrefix+table1).length)); right before the tablesorter line

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread Lukas Pitschl | Dressy Vagabonds
you shouldn't use .length on a jQuery object, since that always returns 1 afaik. Use .size() instead. If your alert then still reports 1 you've selected the table correctly using jQuery, else you have to check your id. also your example implies, that the serverIdPrefix is not used. Check

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread Varun Khatri
*alert($(#+serverIdPrefix+table1).size());// still returns 1* $(#+serverIdPrefix+table1) .tablesorter({widthFixed: true, widgets: ['zebra']}) .tablesorterPager({container: $(#pager)}) *//still doesnot work* Is there any other mistake that you can think of that I am making here , I know

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread MorningZ
you shouldn't use .length on a jQuery object, since that always returns 1 afaik size and length are equivalent, well, except that size is slower Straight from the docs: Length: The number of elements currently matched. The size function will return the same value

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread MorningZ
Can I also use table sorter with datagrid control? No, well, not out of the box anyways... as the DataGrid control doesn't generate thead, th, and tbody tags, all of which are important parts of the TableSorter structure If you insist on using DataGrid, you'll have to make a custom control

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread Varun Khatri
well to me both returned same value So I agree But still cant find solution to problem: *alert($(#+serverIdPrefix+table1).size());// still returns 1* $(#+serverIdPrefix+table1) .tablesorter({widthFixed: true, widgets: ['zebra']}) .tablesorterPager({container: $(#pager)}) *//still

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread MorningZ
Without seeing a working (errr, non working) page, constantly posting the same piece of non-working code is not helping others help you How about just using a class? On Jan 12, 2:51 pm, Varun Khatri khatri.vk1...@gmail.com wrote: well to me both returned same value So I agree But still

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread MorningZ
What happens with: $(.tablesorter) .tablesorter({widthFixed: true, widgets: ['zebra']}) .tablesorterPager({container: $(#pager)}) On Jan 12, 3:11 pm, Varun Khatri khatri.vk1...@gmail.com wrote: Here I have attached 3 files ... Its just a simple example I am trying this coz all my

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread Varun Khatri
If I remove runat=server Like *table id=table1 cellspacing=1 class=tablesorter* And then use : $(.tablesorter) .tablesorter({widthFixed: true, widgets: ['zebra']}) .tablesorterPager({container: $(#pager)}) *It works fine * But If I do *table id=table1 cellspacing=1 class=tablesorter

[jQuery] Re: Jquery tablesorter problem

2009-01-12 Thread MorningZ
Sorry man, no idea what to tell you. good luck with your issue On Jan 12, 3:34 pm, Varun Khatri khatri.vk1...@gmail.com wrote: If I remove runat=server Like *table  id=table1 cellspacing=1 class=tablesorter* And then use : $(.tablesorter)    .tablesorter({widthFixed: true, widgets: