Sql 在同一表格中选择具有不同where条件的命令，该命令在两个parreller列中给出结果

我希望这些结果在平行列中合并 用于登录列

select [Discrepant Reason],count([Discrepant Reason]) as "Login" 
from   Book1 
where ( Remarks<>'Sales Reject' and MONTH([Application Receipt date])=7) 
group by [Discrepant Reason]

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使用条件聚合

select [Discrepant Reason],
sum(case when Remarks='Sales Reject' and MONTH([Application Receipt date])=7 then 1 else 0 end) Login,
sum(case when Remarks<>'Sales Reject' and MONTH([Application Receipt date])=7 then 1 else 0 end) Sales_reject from tble
group by [Discrepant Reason]

选择[差异原因]，
总和（当备注为“销售拒绝”且月份（[申请接收日期]）为7时，则为1，否则为0结束）登录，
金额（备注“销售拒绝”和月份（[申请接收日期]）为7，则为1，否则为0结束）tble拒绝销售
分组依据[差异原因]

这就是您需要的

select [Discrepant Reason],
  sum(case when Remarks='Sales Reject' then 1 else 0 end) Login,
  sum(case when Remarks<>'Sales Reject' then 1 else 0 end) Sales_reject 
from Book1 
where MONTH([Application Receipt date])=7
group by [Discrepant Reason]

选择[差异原因]，
sum（如果备注为“Sales Reject”，则1或0结束）登录，
合计（备注“销售拒绝”时为1，否则为0）销售拒绝
第一册
其中月份（【申请接收日期】=7
分组依据[差异原因]

Hi Zaynul Abadin Tuhin:非常感谢你。你解决了我的问题，谢谢

select [Discrepant Reason],
  sum(case when Remarks='Sales Reject' then 1 else 0 end) Login,
  sum(case when Remarks<>'Sales Reject' then 1 else 0 end) Sales_reject 
from Book1 
where MONTH([Application Receipt date])=7
group by [Discrepant Reason]