PHP Mysql加入问题

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_id association.

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><a href="#">JS</a></li>
                                    <li><a href="#">HTML</a></li>
                                    <li><a href="#">CSS</a></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