递归飞行路径代码VBA语法错误

Question

运行此代码时，会出现语法错误。我无法理解原因
我粘贴了所有的代码。如果你需要，我可以添加数据
我想写递归代码，在时间限制下找到所有可能的死记硬背的航班
find_route函数应在自身中再次运行，直到满足停止条件。（飞行时间>480）

Function find_route(a As Integer, b As Integer) 'a start node, b start time
flight_time = 0
route(0) = a
l = 0
If flight_time <= 480 Then
    If temp_flight_time(a, b) + flight_time <= 480 Then
    l = l + 1
    route(l) = next_destination(a, b)
    flight_time = temp_flight_time(a, b) + flight_time
         find_route(route(l),flight_time)'*******syntax error at this row******
    End If

Else
    Cells(7, 1).Select
    For i = 0 To 30
        ActiveCell.Value = route(i)
        ActiveCell.Offset(0, 1).Select
    Next
    Cells(1, 1).Select
    Exit Function
End If
End Function
Function temp_flight_time(a As Integer, b As Integer)
temp_flight_time = get_flight_time(a, next_destination(a, b))
End Function
Function get_flight_time(a As Integer, b As Integer) 'a from, b to
Cells(2, 1).Select
Dim ae As Integer
Dim be As Integer
For i = 1 To 50
    ae = ActiveCell.Value   'From
    be = ActiveCell.Offset(0, 1).Value  'To
    If a = ae And b = be Then
        get_flight_time = ActiveCell.Offset(0, 4).Value 'Flight Time
        Exit For
    Else
        ActiveCell.Offset(1, 0).Select
    End If
Next
End Function
Function next_destination(a As Integer, b As Integer)
Cells(2, 1).Select
Dim ae As Integer
Dim be As Integer
For i = 1 To 50
    ae = ActiveCell.Value  'To
    be = ActiveCell.Offset(0, 2).Value  'Departure
    If a = ae And b <= be Then
        next_destination = ActiveCell.Offset(0, 1).Value
        Exit For
    Else
        ActiveCell.Offset(1, 0).Select
    End If
Next
End Function

Answer 1

第一行：Function find_route(a As Integer, b As Integer)

find_route是一个函数，因此它返回一个值。未指定返回类型，因此它返回一个Variant。

问题行：find_route(route(l),flight_time)

函数已被调用，但返回值无处可放。这会导致语法错误，因为必须对VBA中的返回值执行某些操作。

由于您从未设置函数的返回值（您可以通过编写find_route = ...来实现），因此您可能希望将其设置为"Sub"。Subs不会返回任何东西，所以这条线现在可以了。

只需将第一行更改为Sub find_route...，并将相应的"End Function"更改为"End Sub"