这个查询也给了我空白的结果！！！ Mysqli 连接表

Question

这张图片显示了一些我不需要它们的 NULL 值的行。

而且这个查询需要很长时间才能加载..

基本上我想要的是

从我的数据库的两个表中获取值时遇到了一些问题。我的数据库中有两个表 1 是 mem，2nd 是支付 mem 存储用户的名称和 drawid 支付表存储用户的提款和分期付款

用户每月向我们付款。因此，如果一个 draw id 为 1 的用户在 2 月向我们付款，则两个表中的值是 mem drawid=1 和 name = something payment draw = 1 和 instalment=2

在mem中的drawid和在payment中的drawid一样

所以表格有多对多的关系。现在我需要找到所有在第 4 个月之前连一期都没有付款的会员列表。

 SELECT drawid,contact,dnd,mem.name,link,address, count(*) as numPayments,NULL numPaidPayments ,NULL PAID_CONTACT,NULL NAME_PAID FROM mem 
LEFT JOIN payment ON (mem.drawid = payment.draw) GROUP BY 
drawid HAVING numPayments < '4'
UNION
SELECT NULL drawid,NULL contact, NULL dnd, NULL name,NULL link, NULL address,NULL numPayments,COUNT(*) as numPaidPayments ,contact PAID_CONTACT,mem.name NAME_PAID  FROM mem 
INNER JOIN payment ON (mem.drawid = payment.draw) GROUP BY 
drawid HAVING numPaidPayments < '4' order by link desc 

我厌倦了解决这个问题，请给我一些建议，我一周以来一直在尝试解决这个问题..

我正在为两个表格结构添加图像。

在此处输入图片描述

Answer 1

您似乎只想计算每个内存的付款：

select 
  m.*,
  (select count(*) from payment p where p.draw = m.drawid) as num_paid_payments
from mem m;

仅限于少于四次付款的人：

select *
from
(
  select 
    m.*,
    (select count(*) from payment p where p.draw = m.drawid) as num_paid_payments
  from mem m
) paid
where num_paid_payments < 4;

或不同的写法：

select 
  m.*,
  coalesce(cnt, 0) as num_paid_payments
from mem m
left join
(
  select draw, count(*) as cnt
  from payment
  group by draw
) p on p.draw = m.drawid
where coalesce(cnt, 0) < 4;

或不同的写法：

select 
  m.*, 
  count(p.draw) as num_paid_payments
from mem m
left join payment p on p.draw = m.drawid
group by m.drawid
having count(p.draw) < 4;