SQL。如果没有相应的正值，则隐藏负值

我有一个SQL语句，如果正数等于负数，它将隐藏记录。但是，如果我意外输入了不正确的负数，则该负数也会在报告中显示为正数，因为我正在使用ABS函数。我不想那样

比如说,， 我的Oracle测试表中有以下布局：

+------+---------+----------+-----------+
| user | project | contract | amount    |
+------+---------+----------+-----------+
| abc  | p123    | c123     |  100.00   |
+------+---------+----------+-----------+
| abc  | p123    | c123     |  200.00   |
+------+---------+----------+-----------+
| abc  | p123    | c123     |  150.00   |
+------+---------+----------+-----------+
| abc  | p123    | c123     | -150.00   |
+------+---------+----------+-----------+

上面的SQL语句将返回前两条记录，这正是我想要的

我遇到的问题是，如果负-150.00错误地输入为-140.00，上面的SQL语句将返回所有四条记录，负-140.00将变为正140.00，这不是我想要的。它应该只显示前三条记录，并隐藏负数-140.00，或任何不具有相等正数的负数

我如何隐藏负数-140.00，只显示前三个记录，并且仍然能够使用GROUP BY和HAVING

---

提前感谢。

您可以按如下方式使用exists：

SELECT user, project, contract, ABS(amount)
FROM test_table t
WHERE contract = 'c123'
AND (amount >= 0 
     or exists (select 1 from test_table tin
                 where t.rowid <> tin.rowid
                   and t.contract = tin.contract 
                   and t.amount + tin.amount = 0)
    )

---

您可以按如下方式使用exists：

SELECT user, project, contract, ABS(amount)
FROM test_table t
WHERE contract = 'c123'
AND (amount >= 0 
     or exists (select 1 from test_table tin
                 where t.rowid <> tin.rowid
                   and t.contract = tin.contract 
                   and t.amount + tin.amount = 0)
    )

---

您可以使用下面的选项获得结果

        WITH data
         AS (SELECT *
             FROM   your table
             WHERE contract = 'c123'
              ),
         d2
         AS (SELECT Row_number()
                      over (
                        PARTITION BY user, project, contract,amount
                        ORDER BY user, project, contract) rw,
                    d.*
             FROM   data d)
    SELECT *
    FROM   (SELECT d.user1,
                   project1,
                   contract,
                   SUM(amount)
                     over(
                       PARTITION BY user, project, contract,Abs(amount), rw
                       ORDER BY user, project, contract) AS amount
            FROM   d2 d)
    WHERE  amount > 0; 

---

您可以使用下面的选项获得结果

        WITH data
         AS (SELECT *
             FROM   your table
             WHERE contract = 'c123'
              ),
         d2
         AS (SELECT Row_number()
                      over (
                        PARTITION BY user, project, contract,amount
                        ORDER BY user, project, contract) rw,
                    d.*
             FROM   data d)
    SELECT *
    FROM   (SELECT d.user1,
                   project1,
                   contract,
                   SUM(amount)
                     over(
                       PARTITION BY user, project, contract,Abs(amount), rw
                       ORDER BY user, project, contract) AS amount
            FROM   d2 d)
    WHERE  amount > 0; 

您可以在不存在的情况下执行此操作：

或者，如果仅合同不足以定义所需的行：

select t.* from test_table t
where t.contract = 'c123' and t.amount > 0
and not exists (
  select 1 from test_table
  where "user" = t."user" and project = t.project 
    and contract = t.contract and amount = -t.amount
)

请参阅。

您可以在不存在的情况下执行此操作：

或者，如果仅合同不足以定义所需的行：

select t.* from test_table t
where t.contract = 'c123' and t.amount > 0
and not exists (
  select 1 from test_table
  where "user" = t."user" and project = t.project 
    and contract = t.contract and amount = -t.amount
)

---

请参阅。

任何天真地使用不存在或类似结构的方法都是危险的。原因很简单：你可以有重复的

因此，我建议使用row_number枚举金额，然后使用：

with tt as (
      select tt.*,
             row_number() over (partition by user, project, contract, amount order by user) as seqnum
      from test_table tt
     ) tt
select tt.*
from tt
where not exists (select 1
                  from tt tt2
                  where tt2.user = tt.user and
                        tt2.project = tt.project and
                        tt2.contract = tt.contract and
                        tt2.seqnum = tt.seqnum and
                        tt2.amount = - tt.amount
                 );

---

任何天真地使用不存在或类似结构的方法都是危险的。原因很简单：你可以有重复的

因此，我建议使用row_number枚举金额，然后使用：

with tt as (
      select tt.*,
             row_number() over (partition by user, project, contract, amount order by user) as seqnum
      from test_table tt
     ) tt
select tt.*
from tt
where not exists (select 1
                  from tt tt2
                  where tt2.user = tt.user and
                        tt2.project = tt.project and
                        tt2.contract = tt.contract and
                        tt2.seqnum = tt.seqnum and
                        tt2.amount = - tt.amount
                 );

---

如果有一个额外的150，而不是，150和-150，它将不会按照您的回答显示。不，它不会显示，因为没有什么可以区分它们。如果有其他列，如可以使用的日期，则OP将澄清。如果有一个额外的150，而不是，150和-150根据您的回答，它不会显示。不会显示，因为没有任何东西可以区分它们。如果有其他列（如日期）可以使用，则OP将澄清。如果有多个金额具有相同的值怎么办？好问题。这是可能的。如果有多个金额具有相同的价值呢？好问题。这是一种可能性。如果我需要显示负数，但没有相应的正数，这很有效。我会保留这个以备将来的需求变更。谢谢！如果我还需要显示负数，但没有相应的正数，那么这个方法很有效。我会保留这个以备将来的需求变更。谢谢！