按子句排序后列值为空

时间：2017-08-28 06:13:01

标签： sql datetime ms-access ms-access-2010 ms-access-2007

如果order by子句未应用于我的列[pay_date]，我将获得列值。下面是创建表和插入虚拟值以设置与我的相同方案的查询

CREATE TABLE [payment_customer]
(
 pay_date datetime ,
 customer_name text,
 pay_amt Memo,
 bal_amt Memo
)


CREATE TABLE [report_invoice]
(
inv_no text,
 inv_date datetime ,
 pname text,
 grand_total Memo
)

Insert Into report_invoice Values 
('INV/17-17/0001',#2017-08-08 00:00:00#,'Customer 1',1000)
Insert Into report_invoice Values 
('INV/17-17/0002',#2017-08-27 00:00:00#,'Customer 1',300)
Insert Into report_invoice Values 
('INV/17-17/0003',#2017-08-27 00:00:00#,'Customer 1',2000)
Insert Into report_invoice Values 
('INV/17-17/0004',#2017-08-27 00:00:00#,'Customer 2',500)
Insert Into report_invoice Values 
('INV/17-17/0005',#2017-08-28 00:00:00#,'Customer 1',3000)
Insert Into report_invoice Values 
('INV/17-17/0006',#2017-08-29 00:00:00#,'Customer 3',700)


Insert Into payment_customer  Values 
(#2017-08-27 00:00:00#,'Customer 1',500,1500)
Insert Into payment_customer  Values 
(#2017-08-28 00:00:00#,'Customer 2',200,300)

最终查询

Select * from
(
Select t1.inv_no,t1.inv_date,t1.pname,t1.grand_total , t2.pay_date,t2.customer_name,t2.pay_amt,t2.bal_amt from report_invoice t1
LEFT  join payment_customer t2 on t1.inv_date = t2.pay_date and t1.pname = t2.customer_name
UNION ALL
Select t2.inv_no,t2.inv_date,t2.pname,t2.grand_total ,  t1.pay_date,t1.customer_name,t1.pay_amt,t1.bal_amt  from payment_customer t1 
LEFT  join report_invoice t2 on  t1.pay_date = t2.inv_date where t2.inv_date is null 
) as v
order by IIF(IsNull(v.inv_date),v.pay_date,v.inv_date)

结果我得到了，但是可以看到customer_name，pay_amt，bal_amt日期为28/08/2017是重复的     inv_no inv_date pname grand_total pay_date customer_name pay_amt bal_amt 

INV/17-18/00001 08/08/2017  Veena Industries Ltd.   238.832 
INV/17-18/00002 27/08/2017  Excel Plants            514.95      27/08/2017        Excel Plants & Equipment Pvt. Ltd.         300    214.95
INV/17-18/00003 29/08/2017  I- Tech                 400 
INV/17-18/00004 28/08/2017  VEENA                   514.95      28/08/2017        VEENA                                      300    214.95
INV/17-18/00005 28/08/2017  VEENA                   600

Result

3 个答案:

答案 0 :(得分：0)

您可能需要指定字段的顺序并在Access SQL语法中使用真正的日期表达式，执行使用货币获取金额：

pay_date datetime,
customer_name text,
pay_amt Currency,
bal_amt Currency

inv_no text,
inv_date datetime,
pname text,
grand_total Currency

Insert Into report_invoice (inv_no,inv_date,pname,grand_total) 
Values ('INV/17-17/0001',#2017-08-08 00:00:00#,'Customer 1',1000)

答案 1 :(得分：0)

经过一些修改后试试这个

Select * from
(
Select t1.inv_no,t1.inv_date,t1.pname,t1.grand_total , t2.pay_date,t2.customer_name,t2.pay_amt,t2.bal_amt from report_invoice t1
LEFT  join payment_customer t2 on t1.inv_date = t2.pay_date and cast(t1.pname as varchar(50)) = cast(t2.customer_name as varchar(50))
UNION ALL
Select t2.inv_no,t2.inv_date,t2.pname,t2.grand_total ,  t1.pay_date,t1.customer_name,t1.pay_amt,t1.bal_amt  from payment_customer t1 
LEFT  join report_invoice t2 on  t1.pay_date = t2.inv_date where t2.inv_date is null 
) as v
order by IIF(v.inv_date is null,v.pay_date,v.inv_date)

Output -

inv_no  inv_date    pname   grand_total pay_date    customer_name   pay_amt bal_amt
INV/17-17/0001  2017-08-08 00:00:00.000 Customer 1  1000.00 NULL    NULL    NULL    NULL
INV/17-17/0002  2017-08-27 00:00:00.000 Customer 1  300.00  2017-08-27 00:00:00.000 Customer 1  500.00  1500.00
INV/17-17/0003  2017-08-27 00:00:00.000 Customer 1  2000.00 2017-08-27 00:00:00.000 Customer 1  500.00  1500.00
INV/17-17/0004  2017-08-27 00:00:00.000 Customer 2  500.00  NULL    NULL    NULL    NULL
INV/17-17/0005  2017-08-28 00:00:00.000 Customer 1  3000.00 NULL    NULL    NULL    NULL
INV/17-17/0006  2017-08-29 00:00:00.000 Customer 3  700.00  NULL    NULL    NULL    NULL

编辑

检查它是否有效

Select v.inv_no,v.inv_date,v.pname,v.grand_total , v.pay_date,v.customer_name,v.pay_amt,v.bal_amt from
(
Select  t1.inv_no,t1.inv_date,t1.pname,t1.grand_total , t2.pay_date,t2.customer_name,t2.pay_amt,t2.bal_amt,ROW_NUMBER() over (partition by t2.pay_amt,t2.bal_amt,t1.inv_date order by t1.inv_date) rn  from report_invoice t1
LEFT  join payment_customer t2 on t1.inv_date = t2.pay_date and cast(t1.pname as varchar(50)) = cast(t2.customer_name as varchar(50))
UNION ALL
Select t2.inv_no,t2.inv_date,t2.pname,t2.grand_total ,  t1.pay_date,t1.customer_name,t1.pay_amt,t1.bal_amt,ROW_NUMBER() over (partition by t1.pay_amt,t1.bal_amt,t2.inv_date order by t2.inv_date) rn  from payment_customer t1 
LEFT  join report_invoice t2 on  t1.pay_date = t2.inv_date where t2.inv_date is null 
) as v
where rn = 1
order by IIF(v.inv_date is null,v.pay_date,v.inv_date)

输出

inv_no  inv_date    pname   grand_total pay_date    customer_name   pay_amt bal_amt
INV/17-17/0001  2017-08-08 00:00:00.000 Customer 1  1000.00 NULL    NULL    NULL    NULL
INV/17-17/0004  2017-08-27 00:00:00.000 Customer 2  500.00  NULL    NULL    NULL    NULL
INV/17-17/0002  2017-08-27 00:00:00.000 Customer 1  300.00  2017-08-27 00:00:00.000 Customer 1  500.00  1500.00
INV/17-17/0005  2017-08-28 00:00:00.000 Customer 1  3000.00 NULL    NULL    NULL    NULL
INV/17-17/0006  2017-08-29 00:00:00.000 Customer 3  700.00  NULL    NULL    NULL    NULL

答案 2 :(得分：0)

如果我的理解是正确的，这就是你需要的。

Select distinct inv_no,inv_date,pname.grand_total,pay_date,customer_name,pay_amt,bal_amt from
(
Select t1.inv_no as inv_no,t1.inv_date as inv_date,t1.pname as pname,t1.grand_total as grand_total , 
t2.pay_date as pay_date,t2.customer_name as customer_name,t2.pay_amt as pay_amt,t2.bal_amt as bal_amt from report_invoice t1
LEFT  join payment_customer t2 on t1.inv_date = t2.pay_date and t1.pname = t2.customer_name
UNION ALL
Select t2.inv_no,t2.inv_date,t2.pname,t2.grand_total ,  t1.pay_date,t1.customer_name,t1.pay_amt,t1.bal_amt  from payment_customer t1 
LEFT  join report_invoice t2 on  t1.pay_date = t2.inv_date where t2.inv_date is null 
) as v
order by IIF(v.inv_date is null,v.pay_date,v.inv_date)

注意： Isnull需要传递两个参数。

CASE可在所有SQL平台上移植，而IIF则是SQL SERVER 2012+特定。

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