根据一个值删除重复项

Question

    customer id name    Pay_type
    1111    aaaa    regular
    1111    aaaa    late
    1111    aaaa    regular
    1111    aaaa    regular
    2222    bbbb    regular
    2222    bbbb    regular
    2222    bbbb    regular
    3333    cccc    regular
    3333    cccc    late
    4444    dddd    regular
    4444    dddd    regular

我有一个 SQL 查询，它给出了上述结果，我希望该结果能删除任何有滞纳金的客户

输出需要是：

customer id name    Pay_type
2222    bbbb    regular
2222    bbbb    regular
2222    bbbb    regular
4444    dddd    regular
4444    dddd    regular

select 
distinct a.customer_id, 
a.name, 
pay_type 
from table a 
left join table b on a.customer_id= b.id 
left join table c on c.id = b.pay_id 
where b.status = 'Done

Answer 1

我不确定你的表格到底是什么样子，但你可以这样做：

WHERE customer_id NOT IN (
    SELECT customer_id
    FROM table_with_customer_and_pay_type
    WHERE pay_type = 'late'
    GROUP BY customer_id )

Answer 2

我会这样做作为反加入：

select *
from table a
where not exists (
  select null
  from table b
  where
    a.customer_id = b.customer_id and
    b.pay_type = 'late'
)

这优于独特的或 "not in" 方法，因为它会在找到匹配项后停止查找。这应该对大型和小型数据集都有效。

任何使用不同的解决方案都必须评估整个数据集，然后删除重复项。

Answer 3

常见Table 表达变体：

WITH orig_result_set AS (
    select 
    distinct a.customer_id, 
    a.name, 
    pay_type 
    from table a 
    left join table b on a.customer_id= b.id 
    left join table c on c.id = b.pay_id 
    where b.status = 'Done'
),

exclude_late_payments AS (
    SELECT DISTINCT customer_id
    FROM orig_result_set
    WHERE pay_type = 'late'
),

on_time_payments AS (
    SELECT customer_id,
           name,
           pay_type
    FROM orig_result_set
    WHERE customer_id NOT IN exclude_late_payments
)

SELECT *
FROM on_time_payments

根据一个值删除重复项

Removing duplicates based on one value

postgresql

duplicates

postgresql-9.4