如何实现数据库查询

时间:2018-10-20 10:17:29

标签: php mysql left-join inner-join

大家好,我应该实现以下可能性:如果包含在另一个表中,则不显示特定值,在这种情况下:仅向某些数据显示未与客户相关联的工厂

"SELECT * FROM store_locator WHERE store_locator.id NOT IN (SELECT impianto_id_campagna FROM campagne_cliente);

但在此查询中:

function get_stores_list2($criteria=array()) {
    $id = $criteria['id'];
    $lat = $criteria['lat'];
    $lng = $criteria['lng'];
    $page_number = $criteria['page_number'];
    $nb_display = $criteria['nb_display'];
    $distance_unit = $criteria['distance_unit'];
    $max_distance = $criteria['max_distance'];
    $category_id = $criteria['category_id'];

    $table_name = $GLOBALS['db_table_name'];
    $start = ($page_number*$nb_display)-$nb_display;

    $s1 = new MySqlTable();

    if($distance_unit=='miles') $distance_unit='3959'; //miles
    else $distance_unit='6371'; //km

    $sql = "SELECT *, 
    ( $distance_unit * acos( cos( radians('".$s1->escape($lat)."') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('".$s1->escape($lng)."') ) + sin( radians('".$s1->escape($lat)."') ) * sin( radians( lat ) ) ) ) AS distance 
    FROM ".$table_name." 
    WHERE 1";

    if($id!='') $sql .= " AND id='".$s1->escape($id)."'";
    if($category_id!='') $sql .= " AND category_id='".$s1->escape($category_id)."'";
    if($max_distance!='') $sql .= " HAVING distance<='".$s1->escape($max_distance)."'";

    if($lat!='' && $lng!='') $sql .= " ORDER BY distance";
    else $sql .= " ORDER BY name";

    if($nb_display>0) $sql .= " LIMIT $start, $nb_display";

    //echo $sql.'<br>';

    $locations = $s1->customQuery($sql);

    return $sql;
}

0 个答案:

没有答案