SQL 根据其他行值排除行答案

【问题标题】：SQL Exclude row based on other rows valueSQL 根据其他行值排除行
【发布时间】：2020-07-23 21:16:46
【问题描述】：

LINE_INVOICE    PAYMENT_SOURCE    SOURCE_PMT_ID                     
-1              Payment Received    7369442                                         
1               Payment Received    7369442                                         
2               Payment Received    7369442                                         
3               Payment Received    7369442                                         
4               Payment Received    7369442                                         
5               Payment Received    7369442                                         
6               Payment Received    7369442                                         
7               Payment Received    7369442                                         
8               Payment Received    7369442                                         
9               Payment Received    7369442                                         
10              Payment Received    7369442                                         
11              Payment Received    7369442                                         
12              Payment Received    7369442

我想在“行发票列”中有任何其他数字时删除“-1”行。这是特定于每个 Source_PMT_ID 的。如果没有'1'行返回-1。如何在我的 sql 查询中捕获它？

【问题讨论】：

在您的示例数据中，您希望删除 LINE_INVOICE 中带有 -1 的行，对吗？

标签： sql oracle where-clause window-functions

【解决方案1】：

我怀疑你想要窗口函数：

select *
from (
    select 
        t.*,
        max(case when line_invoice <> -1 then 1 else 0 end) 
            over(partition by source_pmt_id) flag
    from mytable t
) t
where line_invoice <> -1 or flag = 0

子查询中的窗口max() 检查是否存在至少一行具有相同source_pmt_id 和line_invoice 值而不是-1。外部查询使用该信息过滤掉line_invoice 具有值-1 的行，同时保留那些对于相同source_pmt_id 没有其他值的行。

【讨论】：

Oracle SQL 需要这个
我在 Fiddle 中尝试了这个解决方案，它返回给定数据集的 -1。我认为 OP 要求的是在这种情况下删除 -1。
@LarsSkaug：我的错，最后一个条件必须是flag = 0。固定。
此解决方案在您的代码中将“line”替换为“line_invoice”后也可以使用。

【解决方案2】：

窗口函数的一种方式：

select t.*
from (select t.*, max(line_invoice) over (partition by SOURCE_PMT_ID) as max_line_invoice
      from t
     ) t
where line_invoice > -1 or max_line_invoice = -1;

或者，没有子查询：

select t.*
from t
order by row_number() over (partition by source_pmt_id
                            order by case when line_invoice <> -1 then 1 else 2 end
                           )
fetch first 1 row with ties;

【讨论】：

【解决方案3】：

假设 -1 是最小的数字，并且 line_invoice 从那里上升：

select * 
from table1 tab
where LINE_INVOICE >= case when (select max(line_invoice) 
                                from table1 tb_
                                where tb_.payment_source = tab.payment_source
                                 and tb_.source_pmt_id = tab.source_pmt_id) = -1 then -1 else 0 end

This Fiddle shows how it works.

【讨论】：