Question

I have 2 tables: users and log. Currently my query looks like that.

        $stmt = $this->db->prepare("SELECT u.id, u.email, u.salt, u.pass, u.approved, u.ban, u2.status  FROM `users` AS u LEFT OUTER JOIN `log` AS u2 ON u2.user_id = u.id WHERE u.email = ?") or die($this->db->error);
        $stmt->bind_param("s", $_POST['email']) or die($stmt->error);
        $stmt->execute();
        $stmt->store_result();
        if ($stmt->num_rows == 0) {
            die($this->ajax->respond(7));
        }
        $data = array();
        $stmt->bind_result($data['id'], $data['email'], $data['salt'], $data['pass'], $data['approved'], $data['ban'], $data['status']) or die($stmt->error);
        $stmt->fetch() or die($stmt->error);
        $stmt->close() or die($stmt->error);

Status column of log table – is indicates if user already signed in or not. What I want to do is to check if email exists in users table and to count rows of log table where status=1. Is that possible with one and only query?

In other words:

Here is log table

Take a look at rows where status = 1. This indicates that user 1 currently signed in. To prevent second signin from different browser I want to at first check for email in users table (basic signin procedure) and at that moment check if user not signed in (by counting rows where status = 1 in log table)

Answer 1

Just change the query to this –

SELECT u.id, u.email, u.salt, u.pass, u.approved, 
u.ban, SUM(u2.status) AS status FROM `users` AS u 
LEFT JOIN `log` AS u2 ON u2.user_id = u.id 
WHERE u.email = ?
GROUP BY u.id

There are a number of other ways to do this as well.