比较oracle表中的行并更新匹配的行

Question

我有一张如下表：

**ID      tDate        Product    Price    Quantity    BuySell    Status**
  1     10-May-17       pppp       $12        20         Buy       Null
  2     12-May-17       tttt       $10        20         Sell      Null
  3     12-May-17       tttt       $10        20         Buy       Null
  4     18-May-17       pppp       $14        20         Sell      Null
  5     18-May-17       pppp       $14        20         Buy       Null
  6     18-May-17       pppp       $14        20         Sell      Null

我需要更新名为 STATUS 的字段，并将其设置为“匹配”，只要找到具有相同 tDate、产品、价格和数量且不等于 BuySell 的货币对。

以下是想要的结果：

**ID      tDate        Product    Price    Quantity    BuySell    Status**
  1     10-May-17       pppp       $12        20         Buy       Null
  2     12-May-17       tttt       $10        20         Sell      Matched
  3     12-May-17       tttt       $10        20         Buy       Matched
  4     18-May-17       pppp       $14        20         Sell      Matched
  5     18-May-17       pppp       $14        20         Buy       Matched
  6     18-May-17       pppp       $14        20         Sell      Null

注意#6 是如何不匹配的，因为它只能与另一个 null 匹配。

我希望我可以用一条 SQL 语句来执行此操作。

我现在正在做的可能是最糟糕的方法： 我在 python 中加载到 pandas 数据框，然后循环遍历每一行来比较它们。

s = "SELECT ID, Account, product, Price, tDate, BuySell, Qty" + \
    "FROM Table " + \
    "WHERE Status IS NULL " + \
    "ORDER BY Account, product, tDate, Price, Qty"

df = pd.read_sql(s, conn)

for i in range(len(df.index)-1):

    if df.iloc[i, 1] == df.iloc[i+1, 1]  \
        and df.iloc[i, 2] == df.iloc[i+1, 2] \
        and df.iloc[i, 3] == df.iloc[i+1, 3] \
        and df.iloc[i, 4] == df.iloc[i+1, 4] \
        and df.iloc[i, 5] != df.iloc[i+1, 5] \
        and df.iloc[i, 6] == df.iloc[i+1, 6]:

        s = "UPDATE Temp_Fees " + \
            "SET Strategy = 'UNALLOCATED \ CANCELLED' " + \
            "WHERE ID = " + str(df.iloc[i,0]) + \
            " OR ID = " + str(df.iloc[i + 1, 0])

        #custom function that will execute and commit statement
        bb.EXECUTE(s)

        #avoid reading a matched row 
        i = i + 1

谢谢

Answer 1

merge into mytable t3
using (select t1.*, count(*) over (partition by tdate,product,price,quantity,field) as field2 from
(
select mytable.*, row_number() over (partition by mytable.tdate,mytable.product,mytable.price,mytable.quantity,mytable.buysell 
order by id) field from 
mytable) t1)  t2
on (t2.id=t3.id and t2.field2='2')
when matched then 
update set status='Matched';

Answer 2

这是另一个可以添加到其他观点的观点。这仅解决匹配部分而不是更新或合并部分。我最近遇到了类似的问题，我需要找到与交易日期和地点匹配的记录，但来自两个不同的来源。在这种情况下，记录必须已经排序，以便类似的记录将放在一起。内部查询将记录与之前和之后的记录进行比较，如果它们匹配，则获取它。然后外部查询确定它们是否满足“差异”标准。希望这会有所帮助。

select sbs.trnsid, sbs.amount, sbs.transaction_date, sbs.posted_date, sbs.srcid, 
        sbs.credited_flag, sbs.accid, sbs.compid, sbs.badgeid, sbs.locid, sbs.date_credited, 
        sbs.searchable, sbs.priortime, sbs.nexttime, sbs.priorsource, sbs.nextsource 
    from 
    (select trnsid, amount, transaction_date, posted_date, srcid, credited_flag,
      accid, compid, badgeid, locid, date_credited, transaction_date||locid as searchable,
      lag(transaction_date||locid, 1) over (order by accid) as priortime,
      lead(transaction_date||locid, 1) over (order by accid) as nexttime, 
    lag(srcid, 1) over (order by accid) as priorsource, 
    lead(srcid, 1) over (order by accid) as nextsource
    from transactions_table
    where accid = v_acct
      and transaction_date >= to_date('10/01/2016 00:00:00', 'mm/dd/yyyy hh24:mi:ss') 
      and transaction_date <= to_date('04/23/2017 23:59:59', 'mm/dd/yyyy hh24:mi:ss')
      and srcid in ('B', 'S') order by accid, transaction_date, locid) sbs
    where (sbs.searchable = sbs.nexttime and sbs.srcid = 'S' and sbs.nextsource = 'B')
   or (sbs.searchable = sbs.priortime and sbs.srcid = 'B' and sbs.priorsource = 'S');

Answer 3

您可以获得每个 tdate 的买卖对数量并更新这些行。

MERGE INTO tablename dst
USING (select t.*,count(*) over(partition by tDate,Product,Price,Quantity,rn) as cnt 
       from (select t.*,row_number() over(partition by tDate,Product,Price,Quantity,buysell order by id) as rn
             from tablename t) t
       ) src
ON (src.id = dst.id AND src.cnt=2)
WHEN MATCHED THEN
UPDATE SET Status = 'Matched';

运行此查询以查看如何将行号分配给买卖。

select t.*,count(*) over(partition by tDate,Product,Price,Quantity,rn) as cnt 
from (select t.*,row_number() over(partition by tDate,Product,Price,Quantity,buysell order by id) as rn
      from tablename t) t

Answer 4

未经测试，但仅使用 SQL：

MERGE INTO your_table dst
USING (
  SELECT ROW_NUMBER() OVER (
             PARTITION BY tDate, Product, Price, Quantity, BuySell
             ORDER BY ID
           ) AS idx,
         COUNT( CASE BuySell WHEN 'Buy' THEN 1 END ) OVER (
             PARTITION BY tDate, Product, Price, Quantity
           ) AS num_buy,
         COUNT( CASE BuySell WHEN 'Sell' THEN 1 END ) OVER (
             PARTITION BY tDate, Product, Price, Quantity
           ) AS num_sell
  FROM   your_table
) src
ON ( src.ROWID = dst.ROWID AND src.idx <= LEAST( src.num_buy, src.num_sell ) )
WHEN MATCHED THEN
  UPDATE SET Status = 'Matched';