如果我理解正确的话：你需要做的是找到访客+订单的每个组合的最小日期

select visitor,orders,min(date) as min.date from table group by visitor,orders

这应该是这样的：

visitor orders min.date
  A         0  1-Jan 
  B         0  1-Jan
  B         1  4-Jan
  A         1  12-Jan

这个表（我们称之为tbl）可以自连接以提供

select A.visitor,datediff(day,purchase.date,first.visit) as days.to.purchase 
from (select visitor,min.date as first.visit from tbl where orders=0) A 
inner join (select visitor,min.date as purchase.date from tbl where orders=1) B
on A.visitor=B.visitor

现在，用一个外部查询 Package 此查询，以统计具有相同日期差异的访问者：

select days.to.purchase,count(visitors) as visitors from 
 (select A.visitor,datediff(day,purchase.date,first.visit) as days.to.purchase 
    from (select visitor,min.date as first.visit from tbl where orders=0) A 
    inner join (select visitor,min.date as purchase.date from tbl where orders=1) B
    on A.visitor=B.visitor
) joined
group by days.to.purchase order by days.to.purchase

希望我没听错。我不确定这是不是正确的解决方案，但你没有给我太多的开始：）
完整的解决方案可以是：

select days.to.purchase,count(visitors) as visitors from 
 (select A.visitor,datediff(day,purchase.date,first.visit) as days.to.purchase 
    from 
(select visitor,min.date as first.visit from 
(select visitor,orders,min(date) as min.date from table group by visitor,orders) tbl where orders=0) A 
    inner join 
(select visitor,min.date as purchase.date from 
(select visitor,orders,min(date) as min.date from table group by visitor,orders) tbl where orders=1) B
    on A.visitor=B.visitor
) joined
group by days.to.purchase order by days.to.purchase