Which JOIN to use - php

Question

I'm trying to get info from a table.
I have boards and sub-boards. The column sub_id in table board tells me whether or not the board is a sub-board or not. If it is a sub-board then the id of the parent board is the value, if it isn't then the cell is null.
I wrote this to display whether or not a board has any sub-boards:
<?php
$data = mysql_query("SELECT * FROM board WHERE sub_id='$boardId'")
or die(mysql_error());
while($info = mysql_fetch_array( $data ))
{
Print "<br>Sub-collections:<br>";
Print "<font size='5pt'><a href='/board/index/".$info['id']."'>".$info['board_name']." </a></font>";
}
?>
This works fine.
Now I want to write a similar code to name the parent board (if the board is a sub) but this is where I'm having trouble.
I came up with:
<?php
$data = mysql_query("select * from board WHERE id='$boardId' AND sub_id IS NOT NULL") or die(mysql_error());
$issub = mysql_fetch_array($data);
?>
then using...
<?php if(($issub)): ?>
<?php echo $issub['sub_id'] ?>
<?php else: ?>
<?php endif ?>
To display.
The problem is without a join I cannot show anything other than the id of the parent board. I don't know which columns to join in order to extract info like board_name etc.
EDIT:
popovitsj's suggestion:
<?php
$data = mysql_query("SELECT * FROM board b WHERE EXISTS(SELECT * FROM board WHERE b.id = sub_id)") or die(mysql_error());
while($info = mysql_fetch_array( $data ))
{
Print "<br>Sub-collection of:<br>";
Print "<font size='5pt'><a href='/board/index/".$info['id']."'>".$info['board_name']." </a></font>";
}
?>
This works to a certain extent. It correctly displays the parent board on the subboard page, but also displays it erroneously on the parent board page.
EDIT2:
After researching LEFT JOIN at Meier's suggestion I eventually got it to work using this code:
SELECT * FROM board a LEFT JOIN board b ON a.sub_id = b.id WHERE a.sub_id LIKE b.id AND a.id='$boardId'

Answer 1

Before I answer the question: are you 100% sure that the $boardid does not come directly from http and is therefore manipulable by the user, so your users can not start a sql injection?
The join you need is a "outer self join". Outer join because you also want to display boards that do not have a parent. Outer join is also named left join.
A self join because the table is joined to itself. The trick with the self join is to give the table different names in the statement, so you can distinguish between th parent and the child.
A quick google search for "sql outer self join" gave me this, which explains it quite nicely.
The example is with employees and managers, and the managers are also employees. So the example is quite similar to yours:
http://blog.sqlauthority.com/2010/07/08/sql-server-the-self-join-inner-join-and-outer-join/

Answer 2

There are different solutions to this. One possible one is:
SELECT * FROM Board b
WHERE EXISTS
(SELECT * FROM Board
WHERE b.id = sub_id);
Another possible solution could be this, this only shows 'root' parent boards. (boards with no parents themselves).
SELECT * FROM Board b
WHERE sub_id IS NULL;