Question

找到了很多答案，但仍然无法正常工作，这是我的查询有两个连接

$query = $this->db->select('transaction.*,restaurant.logo_url,restaurant.cover_image_url,restaurant.slug,restaurant.delivery_time,user_points.point AS earned_points')
    ->from('transaction')
    ->join('restaurant', 'transaction.restaurant_id = restaurant.id', 'left')
    ->join('user_points', 'transaction.order_id = user_points.transaction_order_id', 'left')
    ->where_in('order_id', $order_id)
    ->get();
    return ($query->num_rows() < 1) ? null : $query->result();

获取交易表中除'restaurant'和'user_points'表之外的所有记录

连接有什么问题吗？

Answer 1

试试这个

$query = $this->db->select('transaction.*,restaurant.logo_url,restaurant.cover_image_url,restaurant.slug,restaurant.delivery_time,user_points.point AS earned_points',false)
    ->from('transaction')
    ->join('restaurant', 'transaction.restaurant_id = restaurant.id', 'left')
    ->join('user_points', 'transaction.order_id = user_points.transaction_order_id', 'left')
    ->where_in('order_id', $order_id)
    ->get();
    return $query->result();

Answer 2

我强烈建议您使用以下结构。我知道它需要更多行，但更容易理解和维护。

$condition = 'condition';

$t1 = 'table1_name';
$t2 = 'table2_name';
$t3 = 'table3_name';
$get = [
    $t1.'.columna',
    $t1.'.columna',
    $t2.'.columna',
    $t2.'.columna',
    $t3.'.columna',
    $t3.'.columna'
]
$this->db->select($get);
$this->db->from($t1);
$this->db->join($t2, $t1.'.columna = ' . $t2.'.comlumna', 'left');
$this->db->join($t3, $t1.'.columna = ' . $t3.'.comlumna', 'left');
$this->db->where($t1.'.columna', $condition);
$result = $this->db->get()->result();

codeigniter - 如何使用'where_in'子句查询'left join'

2 个答案: