Question

我想问一下如何在for each语句中使用多个break。

样品：

Car Code   Color Code
0001       002
0001       002
0001       001
0005       003
0005       002
0007       001
0008       001
0008       005
0008       001

我的代码是：

def var ctr as int.

对于每辆汽车，通过颜色代码按车号码解锁。

   ctr = ctr + 1.


/*I tried*/
   if last-of(carcode) and last-of(colorcode) then do:
      disp carcode colorcode ctr.
      ctr = 0.
   end. 


/*and tried*/
   last-of(colorcode) then do:
      if last-of(carcode) 
         disp carcode colorcode ctr.
         ctr = 0.
      end.
   end. 
end.

我的预期输出是：

car code   Color Code    QTY
0001       001           1
0001       002           2
0005       002           1
0005       003           1
0007       001           1
0008       001           2
0008       005           1

Answer 1

试试这个：

GHCi> printListTuples [(123, "foo"), (456, "bar")]
"foo 1,2,3\nbar 4,5,6\n"

LAST-OF（颜色代码）可能为真，而LAST-OF（carcode）可能为假，因此将AND更改为OR。

如果LAST-OF（carcode）为真，则LAST-OF（colorcode）也将为真。

Answer 2

当我检查代码时，我忽略了使用last-of并使用了临时表和缓冲区。

def buffer btt-car for tt-car.

find first tt-car where tt-car.carcode = car.carcode exclusive.
if not avail tt-car then do:
  create tt-car.
  assign tt-car.car-code = car.carcode
  tt-car.color-code = car.colorcode
  tt-car.qty = tt-car.qty + car.qty.
end.
if avail tt-car then do:
  find first btt-car where btt-car.colorcode = car.colorcode exclusive.
  if not avail btt-car then do:
    create btt-car.
    assign btt-car.car-code = car.carcode
    btt-car.color-code = car.colorcode
    btt-car.qty = btt-car.qty + car.qty.
  end.
  if avail btt-car then assign btt-car.qty = btt-car.qty + car.qty.
end.

但如果你们有使用最后一个休息时间的解决方案，请分享..

感谢

Answer 3

这样的事情应该有效：

for each car 
    no-lock
    break by car.carcode
          by car.colorcode
:

    accumulate car.colorcode (count by car.colorcode).

    if last-of(car.colorcode)
    then
        display car.carcode 
                car.colorcode
                (accum count by car.colorcode car.colorcode).

end.

当然，如果需要，您可以使用变量而不是ACCUMULATE。

使用Break By for Multiple Fields

3 个答案: