How to enable a tree view in php page? - php

Question

I have a mysql database. In a table i have some entries of uploaded file. I want to display that on a php page. But those files are categorized by one column. So what i want is to have that categories name on that page. In front of each name i want to have a button. Initially when page loads it should not show the files. when some one click on a button then that corresponding category should expand.
But i am unable to do it. I am giving you a link to sample code.
HTML code
<span class="show"></span>
<p>category</p>
<table width='100%' class='tree'>
<tr>
<td width='70%'><b>Name</b></td>
<td width='30%'><b>Last Updated</b></td>
</tr>
</table>
http://jsfiddle.net/SumitRathore/FqcSM/
Edit
this is my sample html code where i am showing the tables. This code have some variables name and sql query don't go for that. I have found the answer here http://jsfiddle.net/JGLaa/2/ but that is not working here. I dont know why?
<span class="show"></span>
<p><b>category<b></p><br/>
<table width='100%' class='tree'>
<tr>
<td width='35%'><b>Name</b></td>
<td width='25%'><b>Last Updated</b></td>
</tr>
while($row = $result_sql->fetch_assoc())
{
<tr>
<td width='50%'><a href='http://127.0.0.1/wordpress/?page_id=464&name_file={$row['name']}&cat={$cat}&sec={$sec}' target='_blank'>{$row['title']}</a></td>
<td width='25%'>{$row['created']}</td>
<td width='25%'><input type='checkbox' class='checkbox'><input type='button' class= 'input1 input_box' value='delete' name='delete' id= '{$row['id']}'><input type='button' class='input2 input_box2' value='edit' name='edit' id= '{$row['id']}'></td>
</tr>
}
</table>

Answer 1

Your issue is that .next() searches the siblings, so it does not find a table. Try traversing up to the buttons parent (span), and then find the span's sibling table -
$(this).parent().nextAll('table:first').find('td').toggle();
updated JSFiddle - http://jsfiddle.net/JGLaa/1/
here is an updated JSFiddle that has multiple buttons/tables - http://jsfiddle.net/JGLaa/2/

Answer 2

Looking at you code the use of the next selector is flawed. From Jquery.com:
.next( [selector ] ) Description:
Get the immediately following sibling of each element in the set of matched elements. If a selector is provided, it retrieves the next sibling only if it matches that selector.
I would suggest to sourround your button and table with a div and use the following code:
$(document).ready(function(){
$(".show").html("<input type='button' value='Show All' class='treebtn'>");
$('.tree td').hide();
$('.treebtn ').click( function() {
$(this).parent().parent().find('table td').toggle();
if (this.value=="Show All") this.value = "Hide All";
else this.value = "Show All";
});
});
http://jsfiddle.net/FqcSM/3/

Answer 3

$(document).ready(function(){
$(".show").html("<input type='button' value='Show All' class='treebtn'>");
$('.tree').hide();
$('.treebtn ').on('click', function() {
$(this).parent('span').next('table').toggle();
if (this.value=="Show All") this.value = "Hide All";
else this.value = "Show All";
});
});
fiddle