i use this table sorterplugin with jquery
http://tablesorter.com/
now my form looks from the structure like this:
<form action="process.php" method="POST">
<table id="banot">
<tr><td>1st line 1 col</td><td> <input type="text" name="i1"/>/td></tr>
<tr><td>2nd line 1 col</td><td> <input type="text" name="i2"/></td></tr>
<tr><td>3rd line 1 col</td><td> <input type="text" name="i3"/></td></tr>
<tr><td>4th line 1 col</td><td> <input type="text" name="i4"/></td></tr>
</table>
<input type="submit" value="submit"/>
</form>
I configured the plugin to seperate all 2 rows a new page. it works perfectly but if i send the form (in my example i have 2 pages), only the actual page’s content is submitted.
Why?
as i understand jquery does hide the other pages only clientside so why the rest of the input don’t get submitted serverside with php?
If i would use tabs then it would work. Whats the difference?
thanks for sheding some light
i searched for a table paginator which only “hides”(still present!) the other pages. Lot of jquery plugins manipulate the whole dom. the problem is that then you can’t use a paged form since on submit all pages/lines have to be present.
My solution was to use a plugin like this:
(the one i’ve chosen)
you can see the difference examining the source with firebug. tablesorter does not have other content present in the table than that is on the current page.