Question

如何选择每个月都有记录的所有客户？在这里，将选择客户，因为他/她每个月都有记录。

+----------+--------+-------+
| customer |  date  | spend |
+----------+--------+-------+
| a        | 201801 |   5.5 |
| b        | 201801 |    16 |
| c        | 201801 |     7 |
| a        | 201802 |   3.2 |
| b        | 201802 |   4.6 |
| a        | 201803 |     3 |
| c        | 201803 |   1.2 |
+----------+--------+-------+

所需的输出：

+----------+--------+-------+
| customer |  date  | spend |
+----------+--------+-------+
| a        | 201801 |   5.5 |
| a        | 201802 |   3.2 |
| a        | 201803 |     3 |
+----------+--------+-------+

Answer 1

我会这样：

  /**
  * @ORM\ManyToMany(targetEntity="Productgroup", inversedBy="fields")
  * @ORM\JoinColumn(name="productgroup", referencedColumnName="id")
  */
  private $productgroup;


public function getProductgroup()
  {
    return $this->productgroup;
  }

  public function setProductgroup($productgroup): self
  {
    $this->productgroup = $productgroup;

    return $this;
  }

  public function __construct()
  {
    $this->productgroup = new ArrayCollection();
  }

Answer 2

假设所有行都有日期为2018，

select customer from customers
group by customer
having count(distinct date) = 12

您可以获取2018年每个月都有行的所有客户。
因此，您需要将其加入主表：

select c.* from (
  select customer from customers
  group by customer
  having count(distinct date) = 12
) t inner join customers c
on c.customer = t.customer
order by c.customer, c.date

如果还有其他年份的日期：

select c.* from (
  select customer from customers
  where substr(date, 1, 4) = '2018'
  group by customer
  having count(distinct date) = 12
) t inner join customers c
where substr(c.date, 1, 4) = '2018'
on c.customer = t.customer
order by c.customer, c.date

Answer 3

假设每个客户每月只有一条记录：

select t.*
from (select t.*, count(*) over (partition by customer) as cnt
      from t
     ) t cross join
     (select count(distinct date) as cnt
      from t
     ) tt
where t.cnt = tt.cnt;

选择所有每月有记录的客户

3 个答案: