vba字典转置字典项目时，得到全零

Question

我要做的第一件事是创建一个脚本字典，将一列中每一行的键设置为零。在随后的代码中，当条件满足时，我将Key的项拉为整数，将其添加1，然后在字典中将0替换为1。

为了测试代码，当我专门调用密钥时，我得到了预期的1。当我将该词典的项目转置为一列时，它给了我全零。当我将其设置为遍历For循环时，将Items放入列中的后续行中，我仍然得到全零。

我不担心最高限额，我正在检查的只有40名招聘人员。

编辑：我睡着了，以为我会放一些更多的测试代码。创建字典后，我立即对字典进行了计数，结果达到了41。在执行完代码后，我又放置了一个计数，发现它达到了42。

什么允许您在字典中拥有两个相同的键？

想法？

LRR和LR？是供参考的最后一行变量

'This creates the dictionary. Recruiter Key with a 0 item
Set RecGoalCount = CreateObject("scripting.dictionary")
    For Each RecDesk In Sheets("Recruiters").Range("B2:B" & LRR)

        RecGoalCount.Add RecDesk, 0

    Next RecDesk
''''Here is the code that will change the dictionary item to a 1. The way _
''''it processes the sheet right now, there is only one instance where all _
''''are true which is why only 1 item gets changed from a 0 to a 1.
RecMatch = 0
For Each job In Sheets("Sheet2").Range("A2:A" & LRS2)

    For Each job2 In Sheets("Sheet1").Range("D2:X" & LRS1)

        OneOfInQW = WorksheetFunction.CountIf(Sheets("Sheet1").Range("D2:X" & LRS1), job)
        CurrentCol = job2.Column
        TrgtCol = CurrentCol - 4
        TrgtCell = job2.Offset(0, -TrgtCol).Value2
        RefCell = job.Offset(0, 1).Value2

            If job = job2 And OneOfInQW = 1 And RefCell >= TrgtCell Then

                    LRS3 = Sheets("Sheet3").Cells(Rows.Count, 1).End(xlUp).Row

                    TrgtRec = CurrentCol - 1
                    TrgtCol = CurrentCol - 2
                    TrgtCell = job2.Offset(0, -TrgtCol).Value2
                    TrgtKey = job2.Offset(0, -TrgtRec).Value2

                    Sheets("Sheet3").Range("A" & (LRS3 + 1)).Value = TrgtCell

                    RecMatch = Int(RecGoalCount(TrgtKey))
                    RecMatch = RecMatch + 1
                    RecGoalCount.Item(TrgtKey) = Str(RecMatch)

                    job2.EntireRow.ClearContents

                    Sheets("Sheet3").Range("B" & (LRS3 + 1)).Value = job
                    Sheets("Sheet3").Range("C" & (LRS3 + 1)).Value = job.Offset(0, 1)

                    job.Clear

                    job.Interior.ColorIndex = 4
                    Sheets("Sheet3").Range("A" & (LRS3 + 1)).Interior.ColorIndex = 4

            Exit For
            End If

    Next job2

Next job
Sheets("Recruiters").Range("E2:E" & LRR) = Application.Transpose(RecGoalCount.Items)
      'This is where I get all 0s despite a condition in the code changing exactly 1 Item to a 1
Sheets("Recruiters").Range("F2:F" & LRR) = Application.Transpose(RecGoalCount.Keys)
      'This is just proving to me that it captured the correct keys with the correct names, check.
Sheets("Recruiters").Range("H2") = RecGoalCount("14CBBPG")
      'This proves to me that the Item became a 1
Recs = RecGoalCount.Count
      'This proves to me that after the code that changes the item, the total _
       count of the dictionary didn't change either.
'This code iterates through the dictionary keys to give the items, and it _
    gives me all 0s despite one of the items being turned into a 1
x = 2
    For Each Rec In RecGoalCount

        Sheets("Recruiters").Range("E" & x) = RecGoalCount(Rec)
        x = x + 1

    Next Rec

我要做的第一件事是创建一个脚本字典，将一列中每一行的键设置为零。在随后的代码中，当条件满足时，我将键的项拉为...

Answer 1

一旦我能够确定自己是在添加新密钥而不是更新密钥，我只需要在Google上搜索字典中可能具有重复外观的密钥。这是我创建字典的方式。我将键添加为单元格而不是值。