Include a php var result on a mysql query - php

Question

It is possible to include PHP data in a MySQL result? Let me explain myself:
Two tables, one with user's actions and one with user information. I'd query the actions and retrieve the user IDs and count each one grouped by user:
$ids = $conn->fetchAll('SELECT origin,COUNT(*) as actions from action WHERE `brand` = ' . $id . ' AND SUBSTRING(origin,1,3)<>"pct" GROUP BY origin');
Then I take that result array and use it to input the user info from another table:
$norm_ids = '(';
foreach ($ids as $ids) {
$norm_ids .= $ids['origin'] .',';
}
$norm_ids = substr_replace($norm_ids ,"",-1) .')';
$users = $conn->fetchAll('SELECT * from userinfo WHERE `id` in ' . $norm_ids . ' ORDER BY `name`');
I want in $users to include the COUNT(*) I got in the previous query, is that possible directly on the query?

Answer 1

I have made my own script, i guess this is what you are looking for:
$sql = "SELECT * FROM user WHERE Username = '".$_SESSION['Username']."' ";
$stm = $db->prepare($sql);
$result = $stm->execute(array());
while($row = $stm->fetch(PDO::FETCH_ASSOC)) {
$userid = $row['UserID'];
}
here i get the users id
$sql = "SELECT activity.*"."FROM user_activity, activity "."WHERE user_activity.ActivityID = activity.ActivityID AND user_activity.UserID = '".$userid."' " ;
and here u use the $userid in my query.

Answer 2

You can use a mysql join for that:
SELECT u.*, count(a.origin) FROM userinfo AS u LEFT JOIN action AS a ON a.origin = u.id WHERE a.brand = '.$id.' AND SUBSTRING(a.origin,1,3) <> "pct" GROUP BY a.origin
You'll have to try and tweak it a little bit (probably) but it might make your script a lot easier. (I'm not sure whether userinfo will be grouped together too or not(probably shouldn't)).
When using a JOIN you'll have to set correct indexes for maximum performance though.

Answer 3

First of all, a few comments on the code in general:
foreach ($ids as $ids)
This just looks wrong, I don't know how PHP handles this but probably not as you would expect it. Use a different variable name, just for clarity's sake.
You need to escape the quotes around "pct" in the query so they're not parsed by PHP.
Also, I don't know where the $id comes from in the first SQL statement, but you might want to escape it. Using PDO:
$stmt = $conn->prepare("SELECT origin,COUNT(*) as actions from action WHERE `brand` = :id AND SUBSTRING(origin,1,3)<>\"pct\" GROUP BY origin");
$stmt->execute( array( ":id" => $id ) );
$ids = stmt->fetchAll();
To access the COUNT(*) data and add it to the user info, I'ld first rename it in the first query (don't like special characters in my array keys):
"SELECT origin,COUNT(*) as nummatches as actions from action WHERE `brand` = :id AND SUBSTRING(origin,1,3)<>\"pct\" GROUP BY origin"
Then, you can add it to the SELECT part of your statement in the second query:
'SELECT *,' . $ids['nummatches'] . ' from userinfo WHERE `id` in ' . $norm_ids . ' ORDER BY `name`'