Flexigrid - Lightweight but rich data grid

[eluser]snobo[/eluser]
Paolo, I'm joining the rest of the crew by saluting you for the fantastic Flexigrid! I seriously wish it becomes THE ultimate grid for us to use. Perhaps with a little help of community...

So, I also found an issue, related to the one noted by apr:

[quote author="apr" date="1213236436"]
if you have more than one flexigrids on one page, the page number and number of pages get copied to all the flexigrids based on the last Ajax request.

I believe the problem is in:

Code:

// around line #547
      $('.pcontrol input').val(p.page);
      $('.pcontrol span').html(p.pages);

[/quote]

The abovementioned problem can be corrected very easily similar to the way it is done a few lines lower (#573), so:

Code:

$('.pcontrol input',this.pDiv).val(p.page);
     $('.pcontrol span',this.pDiv).html(p.pages);

But! I've encountered an interesting situation when having 2 grids on page, one with and another without status bar (pDiv). And how do all the numerous statements like

Code:

$('.pPageStat',this.pDiv).html(p.procmsg);

actually work, when a FlexiGrid has no status, and therefore this.pDiv is void? With no context, they all work by referencing the first .pPageStat on page, wherever it may be located (the other flexiGrid in my case).

So, a straightforward way could be prepending all the statements like one mentioned above, scattered around the flexigrid.js, with if (this.pDiv). However, an easier correction worked well for me. Line #819 originally goes like this:

Code:

if (p.usepager) g.pDiv = document.createElement('div'); //create pager container

I suggest to drop if (p.usepager). So pDiv is always created, even if usepager is set to false. But in that case, it will still contain nothing, so no status bar will be visible.

my 2 cents