PHP Mysql Join issue - php

Question

I am selecting data from 3 tables using a two left joins.
This all works up until the iteration of one of the values for an association (i think, through process of elimination).
I now have reason to believe the select statement is incorrect at the point of the JOINs, but i cannot see how.
The join that is not functioning as expected:
LEFT JOIN
ae_template_pageTypes t ON t.ae_template_pageTypes_id = tp.ae_template_page_group_id
Here i am trying to get the ae_template_pageTypes_type_label by the ae_template_pageTypes_idassociation.
Table: ae_template_pages
Table: ae_template_pageTypes
Table: ae_template_groups
The HTML Output(Both showing HTML instead of using join respectively)
THE ISSUE
I am expecting 2 different values in the format section where here shows both as HTML. I cannot locate why this is happening.
The SELECT
SELECT tg.ae_template_group_name, tp.ae_template_page_id, tp.ae_template_page_group_id, tp.ae_template_page_title, tp.ae_template_page_type, tp.is_group_index, t.ae_template_pageTypes_id, t.ae_template_pageTypes_type_label
FROM
ae_template_pages tp
LEFT JOIN
ae_template_pageTypes t ON t.ae_template_pageTypes_id = tp.ae_template_page_group_id
LEFT JOIN
ae_template_groups tg ON tg.ae_template_group_id = tp.ae_template_page_group_id
WHERE tp.ae_template_page_group_id = '$tempGroup_id'
NOTE: I have tried all types of join to test without finding a solution.
The PHP
foreach ($template_pages as $key => $value) {
?>
<li class="dd-item dd3-item" data-id="<? echo $template_pages[$key]['ae_template_page_id']; ?>">
<div class="dd-handle dd3-handle"></div>
<div class="dd3-content ae_template_page" data-template_page_name='<? echo $template_pages[$key]['ae_template_page_title']; ?>' data-template_page_id='<? echo $template_pages[$key]['ae_template_page_id']; ?>'><? echo $template_pages[$key]['ae_template_page_title']; ?>
<span style='float: right;margin-top: -3px;'>
<div style="" class="btn-group">
<span class="btn-info btn-xs dropdown-toggle" data-toggle="dropdown" title="Click to change file type" type="button"><? echo $template_pages[$key]['ae_template_pageTypes_type_label']; ?></span>
<ul role="menu" class="dropdown-menu" data-page_id='<? echo $template_pages[$key]['ae_template_page_id']; ?>'>
<li>JS</li>
<li>HTML</li>
<li>CSS</li>
</ul>
<span class="btn-info btn-xs" title="This file belongs to Template Group: <? echo $template_pages[$key]['ae_template_group_name']; ?>" type="button"><? echo $template_pages[$key]['ae_template_group_name']; ?></span>
<?
if ($template_pages[$key]['is_group_index'] == 1) {
?>
<span class="btn-success btn-xs" title="This is the default file for this group" type="button">Group index</span>
<?
}
?>
</div>
</span>
</div>
</li>
<?
}
SQL workbench output using the above statement with the $tempGroupId set to 1

Answer 1

For this and future try and restructure your sql to make it clearer eg:
select * from (
(select * from table_a) tbla
left join
(select * from table_b) tblb
on tbla.commoncolname=tblb.commoncolname
)
Try this on the database tool first (eg SQL Workbench), then move on to what the php retrieves and so on to the html