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)
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.
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