分割值有多少？

Question

我有这个字符串

?( {"entity": {"BusinessName": "FANEUIL HALL APARTMENTS, L.L.C.", "UBI": "601754321", "Category": "LLC", "Type": "Profit", "Active": "Active", "StateOfIncorporation": "WA", "DateOfIncorporation": "12/03/1996", "ExpirationDate": "12/31/2014", "DissolutionDate": "", "RegisteredAgentName": "LESLIE R PESTERFIELD", "RegisteredAgentAddress": "901 FIFTH AVE STE 3500", "RegisteredAgentCity": "SEATTLE", "RegisteredAgentState": "WA", "RegisteredAgentZip": "98164", "AlternateAddress": "", "AlternateCity": "", "AlternateState": "", "AlternateZip": "", "GoverningPersons": [ { "Title": "Manager", "LastName": "HEATHMAN", "MiddleName": "W", "FirstName": "MICHAEL W", "Address": "", "City": "PALM SPRINGS", "State": "CA", "Zip": " " } ]}} );

我的目标是让first name，middle name，last name，city和state返回并排列到同一行的不同单元格中但不同的列。

我的代码适用于多个管理人员，但一旦成为一个管理人员，它就会停止按预期工作。

我无法在下面的代码

中计算我的Else部分

Sub test()
    Dim names As String

    Values = Split(resp, """GoverningPersons"":")

    If InStr(1, Values(1), "}, {") > 0 Then
        newvalues = Split(Values(1), "}, {")
        x = Split(newvalues(0), """LastName"": ")
        j = Split(x(1), ", ""MiddleName"":")
        m = Split(j(1), ", ""FirstName"":")
        r = j(0) + m(0) + m(1)
        l = Split(m(1), ", ""Address"": ")
        names = l(0)

        If InStr(1, l(0), "  ") > 0 Then
            al = Left(names, InStr(names, "  ") - 1)
            d = al + m(0) + j(0)
        Else
            d = l(0) + m(0) + j(0)
        End If

        a = Replace(d, Chr(32), "")
        g = Replace(a, Chr(34), " ")
    ElseIf InStr(1, Values(1), "[ }}") > 0 Then

    Else

    End If
End Sub

我尝试了所有我能想到的东西，这个名字永远不会像我代码的第一部分那样出现。

Answer 1

这将帮助您入门。

我在这里向您展示了如何提取First name，middle name和last name。我相信您可以应用相同的逻辑来提取city和state。

注意：我没有包含任何错误处理。我相信你会照顾好吗？

Sub Sample()
    Dim resp As String
    Dim Lastname As String, Midname As String, Firstname As String

    '~~> For testing purpose, I stored the json string in Cell A1
    resp = Range("A1").Value

    '~~> This will give you last name
    Lastname = GetValue(resp, """LastName"":")
    Debug.Print "Lastname :" & Lastname

    '~~> This will give you middle name
    Midname = GetValue(resp, """MiddleName"":")
    Debug.Print "Middle Name :" & Midname

    '~~> This will give you First name
    Firstname = GetValue(resp, """FirstName"":")
    Debug.Print "First Name :" & Firstname
End Sub

Function GetValue(sString As String, sDelim As String) As String
    Dim MyArray, tmpAr

    MyArray = Split(sString, sDelim)
    tmpAr = Split(MyArray(1), ",")
    GetValue = Trim(Replace(tmpAr(0), Chr(34), ""))
End Function

<强>截图

Excel表格

VBA编辑

Answer 2

这是另一种方法（需要一些javascript / JSON知识）

Sub tester()

'add project reference to "Microsoft script control"
Dim so As New ScriptControl
Dim i As Integer, num As Integer
Dim js As String, json As String

    so.Language = "jscript"


    json = "{""entity"": {""BusinessName"": ""FANEUIL HALL APARTMENTS, L.L.C."", " & _
        " ""UBI"": ""601754321"", ""Category"": ""LLC"", ""Type"": ""Profit"", ""Active"": " & _
        " ""Active"", ""StateOfIncorporation"": ""WA"", ""DateOfIncorporation"": " & _
        " ""12/03/1996"", ""ExpirationDate"": ""12/31/2014"", ""DissolutionDate"": """", " & _
        " ""RegisteredAgentName"": ""LESLIE R PESTERFIELD"", ""RegisteredAgentAddress"": " & _
        " ""901 FIFTH AVE STE 3500"", ""RegisteredAgentCity"": ""SEATTLE"", " & _
        " ""RegisteredAgentState"": ""WA"", ""RegisteredAgentZip"": ""98164"", " & _
        " ""AlternateAddress"": """", ""AlternateCity"": """", ""AlternateState"": """", " & _
        " ""AlternateZip"": """", ""GoverningPersons"": [  { ""Title"": ""Manager"", " & _
        " ""LastName"": ""HEATHMAN"", ""MiddleName"": ""W"", ""FirstName"": ""MICHAEL W"", " & _
        " ""Address"": """", ""City"": ""PALM SPRINGS"", ""State"": ""CA"", ""Zip"": """" } ]}}"

     so.AddCode "var o = eval(" & json & ");"

     num = so.Eval("o.entity.GoverningPersons.length") ' >> 1

     For i = 0 To num - 1
        js = "o.entity.GoverningPersons[" & i & "]."
        Debug.Print so.Eval(js & "LastName;")
        Debug.Print so.Eval(js & "FirstName;")
        Debug.Print so.Eval(js & "MiddleName;")
        'etc
     Next i


End Sub