How do I ignore results from JOIN table in MySQL query

Question

I have the following query:

$img_sub_qry = "SELECT * FROM images 
JOIN imagsub ON images.image_id = imagsub.image_id
WHERE images.image_id IN ($thelist)";

The problem is that this query is populating a drop down box and I am getting double results. I only want the results from the images table but I need the JOIN in order to grab a column that only exists in the imagsub table. From FirePHP I can see the results from the images table coming through just fine but then it loops back and pulls the same results from imagsub and I don't need those results. I have tried "SELECT DISTINCT" to no avail. Can someone help me here. I am fairly new to PHP and this is the first time I have tried using IN.

Answer 1

I think you're on the right track with select distinct; just specify the columns you want. If you're populating a drop-down list, go like this:

$img_sub_qry = "SELECT DISTINCT [colname] as ddlvalue, [colname] as ddltext
    FROM images 
    JOIN imagsub ON images.image_id = imagsub.image_id
    WHERE images.image_id IN ($thelist)";

For the query you're trying, every row in the query is going to be distinct, probably, because of the ID columns.

Answer 2

You should explicitly specify what fields you are interested in.

SELECT i.* 
FROM images i
JOIN imagsub is ON i.image_id = is.image_id
WHERE i.image_id IN ($thelist)