Question

这里我想得到两个表值的总和，我的表名是datas和abc_datas。首先，我获取datas表的结果如下 -

$query1 = $this->db->query("SELECT *, SUM(amount) AS total_sum
         FROM datas
         WHERE user_id = $user_id")->row();

接下来，我按如下方式获取abc_datas的结果 -

$query2 = $this->db->query("SELECT *, SUM(amount) AS total_sum
         FROM abc_datas
         WHERE user_id = $user_id")->row();

现在，我希望将这两个查询组合在一起，并通过组合total_sum来返回总数。例如，如果第一个表的total_sum是500而第二个表是1000，我希望输出为1500，但只有第一个查询值被取出

这是我写的最后一个查询 -

$query = $this->db->query("SELECT *, SUM(amount) AS total_sum1
     FROM datas
     WHERE user_id = $user_id
     UNION ALL
     SELECT *, SUM(amount) AS total_sum2
     FROM abc_datas
     WHERE user_id = $user_id;")->row();
    return $query;

我改变了我的查询，以获得所需的输出 -

 $query =$this->db->query("SELECT
 (SELECT SUM(amount) as total_sum1 FROM datas WHERE    user_id=$user_id) + (SELECT SUM(amount) as total_sum2 FROM abc_datas WHERE user_id=$user_id) as total_sum,(SELECT SUM(agent_amount) as total_agent_amount1 FROM datas WHERE    user_id=$user_id) + (SELECT SUM(agent_amount) as total_agent_amount2 FROM abc_datas WHERE user_id=$user_id) as total_agent_amount,(SELECT SUM(profit) as total_profit1 FROM datas WHERE    user_id=$user_id) + (SELECT SUM(profit) as total_profit2 FROM abc_datas WHERE user_id=$user_id) as total_profit
 FROM dual");
 return $query->row();

Answer 1

结果如下

 $query = $this->db->query("SELECT SUM(amount) AS total_sum
        FROM   (SELECT   amount FROM datas WHERE    `user_id`=$user_id
                   UNION ALL
                SELECT amount FROM abc_datas WHERE    `user_id`=$user_id) s");
   return $query->row();

我们可以组合两个查询返回结果为行

1 个答案: