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POQuery with sub-results from another table
 

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    <p>I have the following two database tables:</p> <pre><code>`membership_members` ( `id` int(11) NOT NULL AUTO_INCREMENT, `name` tinytext NOT NULL, `active` tinyint(1) NOT NULL DEFAULT '1', PRIMARY KEY (`id`) ) `membership_payments` ( `date` date NOT NULL, `amount` int(11) NOT NULL, `member` int(11) NOT NULL, KEY `member` (`member`) ) </code></pre> <p>I wish to list all the members, and for each of them I would like to show the list of all the payments by that member. The <code>membership_payments.member</code> column is a foreign key <code>membership_members.id</code> (this is MySQL so I cannot explicitly specify a foreign key). Notice that I do want to list a member even if he doesn't have any payments.</p> <p>So something like:</p> <pre><code>* John Smith - 2012-05-06 $100 - 2012-01-02 $100 * Brian Jones * Mike Jackson - 2012-09-02 $50 </code></pre> <p>I have tried with this query:</p> <pre><code>SELECT id, name, active, date, amount FROM `membership_members`, `membership_payments` WHERE membership_members.id = member </code></pre> <p>This of course gives me tha data I need, but not exactly how I need it as it returns a row for each payment. That means that I later have to group the rows (in PHP) by member id so that I don't list a member multiple times. I <em>can</em> do that, but I believe that it would be easier and faster to do it in SQL. Also, this query only gives me users which have payments with their id.</p> <p>I feel that this should be simple, but last time I did anything but the most simple stuff in SQL was 6-7 years ago.</p> <p><strong>EDIT:</strong> <code>LEFT OUTER JOIN</code> suggested in one of the answers solves the issue with the "missing" members. But what is the best way of grouping results by member IDs? I know there is a <code>GROUP BY</code> clause, but it doesn't give me all the payments for the given member, only one. I suppose I can run a new payments query for each member, but I fear this would be very inefficient (we have around 300-400 members).</p>
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