Sql 如何将状态添加到表中
我有下表，其中是从我的数据库剪辑。我有两种合同 I:客户支付前6个月60美元，下一个6个月120美元111客户 II：客户支付前6个月的60美元，但如果仍然支付60美元，合同将在第6个月延长，整个合同为18个月。321名仍在付款的客户Sql 如何将状态添加到表中,sql,oracle,plsql,Sql,Oracle,Plsql,我有下表，其中是从我的数据库剪辑。我有两种合同 I:客户支付前6个月60美元，下一个6个月120美元111客户 II：客户支付前6个月的60美元，但如果仍然支付60美元，合同将在第6个月延长，整个合同为18个月。321名仍在付款的客户 ID_Client | Amount | Amount_charge | Lenght | Date_from | Date_to | Reverse ---------------------------------------------------
ID_Client | Amount | Amount_charge | Lenght | Date_from | Date_to | Reverse
--------------------------------------------------------------------------------
111 60 60 12 2015-01-01 2015-01-31 12
111 60 60 12 2015-02-01 2015-02-28 11
111 60 60 12 2015-03-01 2015-03-31 10
111 60 60 12 2015-04-01 2015-04-30 9
111 60 60 12 2015-05-01 2015-05-31 8
111 60 60 12 2015-06-01 2015-06-30 7
111 120 60 12 2015-07-01 2015-07-31 6
111 120 60 12 2015-08-01 2015-08-31 5
111 120 60 12 2015-09-01 2015-09-30 4
111 120 60 12 2015-10-01 2015-10-31 3
111 120 60 12 2015-11-01 2015-11-30 2
111 120 60 12 2015-12-01 2015-12-31 1
111 120 60 12 2016-01-01 2015-01-31 0
111 120 60 12 2016-02-01 2015-02-29 0
321 60 60 12 2015-01-01 2015-01-31 12
321 60 60 12 2015-02-01 2015-02-28 11
321 60 60 12 2015-03-01 2015-03-31 10
321 60 60 12 2015-04-01 2015-04-30 9
321 60 60 12 2015-05-01 2015-05-31 8
321 60 60 12 2015-06-01 2015-06-30 7
321 60 60 12 2015-07-01 2015-07-31 6
321 60 60 12 2015-08-01 2015-08-31 5
321 60 60 12 2015-09-01 2015-09-30 4
321 60 60 12 2015-10-01 2015-10-31 3
321 60 60 12 2015-11-01 2015-11-30 2
321 60 60 12 2015-12-01 2015-12-31 1
321 60 60 12 2016-01-01 2016-01-30 0
321 60 60 12 2016-02-01 2016-02-31 0
321 60 60 12 2016-03-01 2016-03-30 0
321 60 60 12 2016-04-01 2016-04-31 0
ID_Client | Amount | Amount_charge | Lenght | Date_from | Date_to | Reverse | Status
-----------------------------------------------------------------------------------------
111 60 60 12 2015-01-01 2015-01-31 12 A
111 60 60 12 2015-02-01 2015-02-28 11 A
111 60 60 12 2015-03-01 2015-03-31 10 A
111 60 60 12 2015-04-01 2015-04-30 9 A
111 60 60 12 2015-05-01 2015-05-31 8 A
111 60 60 12 2015-06-01 2015-06-30 7 A
111 120 60 12 2015-07-01 2015-07-31 6 D
111 120 60 12 2015-08-01 2015-08-31 5 D
111 120 60 12 2015-09-01 2015-09-30 4 D
111 120 60 12 2015-10-01 2015-10-31 3 D
111 120 60 12 2015-11-01 2015-11-30 2 D
111 120 60 12 2015-12-01 2015-12-31 1 D
111 120 60 12 2016-01-01 2015-01-31 0 E
111 120 60 12 2016-02-01 2015-02-29 0 E
321 60 60 12 2015-01-01 2015-01-31 12 A
321 60 60 12 2015-02-01 2015-02-28 11 A
321 60 60 12 2015-03-01 2015-03-31 10 A
321 60 60 12 2015-04-01 2015-04-30 9 A
321 60 60 12 2015-05-01 2015-05-31 8 A
321 60 60 12 2015-06-01 2015-06-30 7 A
321 60 60 12 2015-07-01 2015-07-31 6 L
321 60 60 12 2015-08-01 2015-08-31 5 L
321 60 60 12 2015-09-01 2015-09-30 4 L
321 60 60 12 2015-10-01 2015-10-31 3 L
321 60 60 12 2015-11-01 2015-11-30 2 L
321 60 60 12 2015-12-01 2015-12-31 1 L
321 60 60 18 2016-01-01 2016-01-30 0 L
321 60 60 18 2016-02-01 2016-02-31 0 L
321 60 60 18 2016-03-01 2016-03-30 0 L
321 60 60 18 2016-04-01 2016-04-31 0 L
如果反向列是我所想的：
update table1 a
set "Status"=
CASE
WHEN A."Reverse" > 6 THEN
'A'
WHEN A."Reverse" > 0 THEN
DECODE (A."Amount", A."Amount_charge", 'L', 'D')
ELSE
CASE
WHEN A."Amount" <> A."Amount_charge" THEN
'E'
ELSE
CASE WHEN ADD_MONTHS ( (SELECT b."Date_from" FROM table1 b WHERE a."ID_Client" = b."ID_Client" AND b."Reverse" = 1),6) > a."Date_from" THEN 'L'
ELSE
'E'
END
END
END
DECLARE
CURSOR c2
IS
SELECT ID_CLIENT, --AMOUNT, AMOUNT_CHARGE, LENGTH, DATE_FROM, DATE_TO, REVERSE, STATUS,
FIRST_VALUE (amount_charge) OVER (PARTITION BY id_client ORDER BY date_from) first_amount_charge,
SUM (amount) OVER (PARTITION BY id_client ORDER BY date_from) sum_amount,
SUM (amount_charge) OVER (PARTITION BY id_client ORDER BY date_from) sum_amount_charge
FROM TABLE2
FOR UPDATE NOWAIT;
BEGIN
FOR c1 IN c2
LOOP
UPDATE table2
SET status = CASE WHEN c1.sum_amount <= 6 * c1.first_amount_charge THEN 'A'
WHEN c1.sum_amount > 18 * c1.first_amount_charge THEN 'E'
WHEN c1.sum_amount > c1.sum_amount_charge THEN 'D'
ELSE 'L'
END
WHERE CURRENT OF c2;
END LOOP;
END;
什么是相反的。倒档是否总是从12开始，并且不能小于0？只有60美元和120美元的矿石还有其他的吗？@motor:这取决于长度栏。反向计数从长度的第一个值开始。您是否知道如何区分客户在6个月后的金额是否翻倍？金额可能不同，例如20美元、30美元、50美元等等。您看到答案了吗。如果不想比较总和，可以使用分析函数ROW_NUMBER OVER。。。