Question

我目前在SQL Server 2012中有以下查询。除了'cd.billed_amt as [Disb Billed]'之外的字段将返回一行，这可以为单个bill_num提供多个结果。

我想要做的只是汇总[Disb Billed]列，以便初始的8行结果集将成为2行结果集。

select distinct bbl.bill_num, 
       bb.tran_type, 
       hm.clnt_matt_code, 
       bb.tran_date, 
       bb.period, 
       cd.billed_amt as [Disb Billed], 
       fees_amt
  from blt_bill_amt bb
  join hbm_matter hm on bb.matter_uno = hm.matter_uno
  join blt_billm bbm on bbm.billm_uno = bb.billm_uno
  join blt_bill bbl on bbl.tran_uno = bbm.bill_tran_uno
  left outer join cdt_disb cd on cd.bill_tran_uno = bbl.tran_uno
 where bb.tran_type in ('WO', 'WOX') 
  and bb.period = '201401'
  and bbl.bill_num = 231728
order by  bb.tran_type, bbl.bill_num

当前结果集

bill_num tran_type clnt_matt_code       tran_date                period   Disb Billed   fees_amt
------------------------------------------------------------------------------------------------
231728   WO         N10118.1016         2013-04-18 00:00:00.000  201401   3.00          8.06
231728   WO         N10118.1016         2013-04-18 00:00:00.000  201401   20.00         8.06
231728   WO         N10118.1016         2013-04-18 00:00:00.000  201401   38.00         8.06
231728   WO         N10118.1016         2013-04-18 00:00:00.000  201401   42.50         8.06
231728   WO         N10118.1016-0001    2013-04-18 00:00:00.000  201401   3.00          0.94
231728   WO         N10118.1016-0001    2013-04-18 00:00:00.000  201401   20.00         0.94
231728   WO         N10118.1016-0001    2013-04-18 00:00:00.000  201401   38.00         0.94
231728   WO         N10118.1016-0001    2013-04-18 00:00:00.000  201401   42.50         0.94

期望的结果集

bill_num tran_type clnt_matt_code       tran_date                period   Disb Billed   fees_amt
------------------------------------------------------------------------------------------------
231728   WO         N10118.1016         2013-04-18 00:00:00.000  201401   103.50        8.06
231728   WO         N10118.1016-0001    2013-04-18 00:00:00.000  201401   103.50        0.94

Answer 1

您需要使用GROUP BY reference

对结果进行分组

所以对你的例子来说：

select distinct bbl.bill_num, 
       bb.tran_type, 
       hm.clnt_matt_code, 
       bb.tran_date, 
       bb.period, 
       SUM(cd.billed_amt) as [Disb Billed], --sum the billed column
       fees_amt
  from blt_bill_amt bb
  join hbm_matter hm on bb.matter_uno = hm.matter_uno
  join blt_billm bbm on bbm.billm_uno = bb.billm_uno
  join blt_bill bbl on bbl.tran_uno = bbm.bill_tran_uno
  left outer join cdt_disb cd on cd.bill_tran_uno = bbl.tran_uno
 where bb.tran_type in ('WO', 'WOX') 
  and bb.period = '201401'
  and bbl.bill_num = 231728
GROUP BY bbl.bill_num, --add group by
       bb.tran_type, 
       hm.clnt_matt_code, 
       bb.tran_date, 
       bb.period,
       fees_amt
order by  bb.tran_type, bbl.bill_num

SQL - 汇总结果集中的单个列

1 个答案: