Question

您如何在bartender vb脚本中实现以下逻辑？

while ( ReceiveNextEvent ( 0, NULL, 0.001, // kEventDurationForever, kEventRemoveFromQueue, &theEvent ) == noErr ) { if (GetEventKind(theEvent) == kEventAppleEvent) AEProcessEvent(theEvent); SendEventToEventTarget(theEvent, theTarget); ReleaseEvent(theEvent); } = 33 + 1 + 8 + 38 + 31 + 21 + 4 + 0 + 1 + 9 + 7 + 3 + 6 + 6 = 168 X18 VL40197366，X=33，" "=38。

现在上面的结果存储到名为“temp1”的变量中其中temp1包含now = 168; `temp1中= 168``

我的代码如下。我知道我的代码不完整，如果我错了请纠正我

V=31, L=21

如何实现这一点！

提前致谢

Answer 1

首先，我显然不明白你想要做什么。

但是，在您的情况下，您应该使用Dictionary类而不是声明单个变量。

Dim dicResult as new Dictionary(Of String, Decimal)
Dim arrSymballs() As String = {"A","B","C","D","E","F","G","H","I"} ''You can add more
Dim iValue As Decimal  = 10
For i As Integer = 0 To arrSymballs.Length-1
    dicResult.Add(arrSymballs(i), iValue)
    iValue += 1
Next

''Function to get string from result
''I don't know how the result you will get the following logic is for  "10,11,12,13,14,15"
Private Function getResultString(_result As String) As Decimal
    Dim dNumResult As Decimal
    For Each c As String in _result.Split(",")
        dNumResult += dicResult(c)
    Next
    Return dNumResult
End Function

Answer 2

您没有将变量声明为数字类型。编译器猜测您希望它们是字符串，因为Format返回一个字符串。您必须将字符串值转换为数值。

Integer.Parse（文本）

temp6 = Integer.Parse(temp5) + x

当您拥有大量变量时，请考虑使用数组。

Dim iValues As Integer() = New Integer() {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35}

temp6 = Integer.Parse（temp5）+ iValues（24）

如何在bartender标签vb脚本中添加字符串中的数字

2 个答案: