从多个表中选择过滤器

Question

我有一个包含2个表的用户聊天系统：users表和followers表。 我正在执行搜索，我想吸引我关注的用户

users表具有此键作为主键： user_id和followers表具有以下列following_id follower_id is_typing active

我如何加入两个表以获得结果？

我一直在这样做：

function Wo_GetNearbyFriends($args = array()) {
    global $wo, $sqlConnect;
    if ($wo['loggedin'] == false || empty($args)) {
        return false;
    }

    $options      = array(
        "offset" => false,
        "gender" => false,
        "name" => false,
        "distance" => false,
        "relship" => false,
        "status" => false,
        "fid_5" => false,
        "fid_6" => false,
        "fid_7" => false,
        "fid_8" => false,
        "limit" => 20
    );
    $args         = array_merge($options, $args);
    $offset       = Wo_Secure($args['offset']);
    $gender       = Wo_Secure($args['gender']);
    $name         = Wo_Secure($args['name']);
    $loc_distance = Wo_Secure($args['distance']);
    $status       = Wo_Secure($args['status']);
    $relship      = Wo_Secure($args['relship']);
    $fid_5      = Wo_Secure($args['fid_5']);
    $fid_6      = Wo_Secure($args['fid_6']);
    $fid_7      = Wo_Secure($args['fid_7']);
    $fid_8      = Wo_Secure($args['fid_8']);
    $limit        = Wo_Secure($args['limit']);
    $unit         = 6371;
    $user_lat     = $wo['user']['lat'];
    $user_lng     = $wo['user']['lng'];
    $user         = $wo['user']['id'];
    $t_users      = T_USERS;
    $t_followers  = T_FOLLOWERS;
    $distance     = 25;
    $data         = array();
    $sub_sql      = "";
    if ($loc_distance && is_numeric($loc_distance) && $loc_distance > 0) {
        $distance = $loc_distance;
    }
    if ($name) {
        $name = Wo_Secure($name);
        $sub_sql .= " AND (`username` LIKE '%$name%' OR `first_name` LIKE '%$name%' OR `last_name` LIKE '%$name%') ";
    }
    if (isset($status) && $status != false) {
        if ($status == 1) {
            $time = time() - 60;
            $sub_sql .= " AND `lastseen` > '$time'";
        } else if ($status == 0) {
            $time = time() - 60;
            $sub_sql .= " AND `lastseen` < '$time'";
        }
    }
    if ($relship && in_array($relship, array_keys($wo['relationship']))) {
        $sub_sql .= " AND `relationship_id`  = '$relship' ";
    }
    if ($offset && is_numeric($offset) && $offset > 0) {
        $sub_sql .= " AND `user_id` <  '$offset' AND `user_id` <> '$offset' ";
    }
    if ($gender && in_array($gender, array_keys($wo['genders']))) {
        $sub_sql .= " AND `gender` = '$gender' ";
    }
    if($fid_5 && is_numeric($fid_5) && $fid_5 > 0){
        $sub_sql .= " AND `fid_5` = '$fid_5' ";
    }
    if($fid_6 && is_numeric($fid_6) && $fid_6 > 0){
        $sub_sql .= " AND `fid_6` = '$fid_6' ";
    }
    if($fid_7 && is_numeric($fid_7) && $fid_7 > 0){
        $sub_sql .= " AND `fid_7` = '$fid_7' ";
    }
    if($fid_8 && is_numeric($fid_8) && $fid_8 > 0){
        $sub_sql .= " AND `fid_8` = '$fid_8' ";
    }

    $sql   = "SELECT `user_id`, ( {$unit} * acos(cos(radians('$user_lat')) *  cos(radians(lat)) * cos(radians(lng) - radians('$user_lng')) +  sin(radians('$user_lat')) * sin(radians(lat))) ) AS distance FROM $t_users WHERE `user_id` <> '$user' {$sub_sql}  AND `user_id` IN (SELECT `follower_id` FROM $t_followers WHERE `follower_id` <> {$user} AND `following_id` = {$user} AND `active` = '1') AND `user_id` IN (SELECT `following_id` FROM $t_followers WHERE `follower_id` = {$user} AND `following_id` <> {$user} AND `active` = '1') AND `lat` <> 0 AND `lng` <> 0  HAVING distance < '$distance' ORDER BY `user_id` DESC LIMIT 0, $limit ";

    $query = mysqli_query($sqlConnect, $sql);

    while ($fetched_data = mysqli_fetch_assoc($query)) {
        $fetched_data['user_data']        = Wo_UserData($fetched_data['user_id']);
        $fetched_data['user_data']['age'] = Wo_GetUserCountryName($fetched_data['user_data']);
        $fetched_data['user_geoinfo']     = $fetched_data['user_data']['lat'] . ',' . $fetched_data['user_data']['lng'];
        if ($fetched_data['user_data']['share_my_location'] == 1) {
            $data[] = $fetched_data;
        }
    }
    return $data;
}

Answer 1

这仅仅是联接。 一旦有了两个用户行，就可以将距离计算应用于他们的数据。

select *
from t_users u1
join t_users u2
  on u1.user_id < u2.user_id
  and (u1.user_id, u2.user_id) in (select follower_id, following_id from t_followers)
  and (u1.user_id, u2.user_id) in (select following_id, follower_id from t_followers);

演示： https : //dbfiddle.uk/?rdbms=mysql_8.0&fiddle=f29e1d27055575385d4b28d777fd335c