有一个程序可以正常工作。她的工作结果是Excel中元素表(href)的输出(每个元素看起来像:about:new_ftour.php?champ = 2604& f_team = 412& tour = 110)。我想用超链接替换href(用“http://allscores.ru/soccer/”替换文本“about:”)。在一行(oRange.Value = data)之后我添加了一行(oRange.Replace What:=“about:”,Replacement:=“http://allscores.ru/soccer/”)。但由于神秘的原因,程序会出错(运行时错误'91')。在行中(Loop While Not R is Nothing and r.Address<> firstAddress and iLoop< 19)。
Sub Softгиперссылки()
Application.DisplayAlerts = False
Call mainмассивы
Application.DisplayAlerts = True
End Sub
Sub mainмассивы()
Dim r As Range
Dim firstAddress As String
Dim iLoop As Long
Dim book1 As Workbook
Dim sheetNames(1 To 19) As String
Dim Ssilka As String
sheetNames(1) = "Лист1"
sheetNames(2) = "Лист2"
sheetNames(3) = "Лист3"
sheetNames(4) = "Лист4"
sheetNames(5) = "Лист5"
sheetNames(6) = "Лист6"
sheetNames(7) = "Лист7"
sheetNames(8) = "Лист8"
sheetNames(9) = "Лист9"
sheetNames(10) = "Лист10"
sheetNames(11) = "Лист11"
sheetNames(12) = "Лист12"
sheetNames(13) = "Лист13"
sheetNames(14) = "Лист14"
sheetNames(15) = "Лист15"
sheetNames(16) = "Лист16"
sheetNames(17) = "Лист17"
sheetNames(18) = "Лист18"
sheetNames(19) = "Лист19"
'пропускаем ошибку
Set book1 = Workbooks.Open("E:\Super M\Проект ставки\Поиск решения\Усов 7\Условия для андердогов\пробная.xlsm")
iLoop = 0
With book1.Worksheets("Лист1").Range("S34:S99") '<--| open wanted workbook and refer to cells "U33:U99" in its worksheet "7"
Set r = .Find(What:="1", LookIn:=xlValues) '<--| the Find() method is called on the range referred to in the preceding With statement
If Not r Is Nothing Then
firstAddress = r.Address
Do
iLoop = iLoop + 1
Ssilka = r.Offset(, -14).Hyperlinks.Item(1).Address
.Parent.Parent.Worksheets(sheetNames(1)).Activate
.Parent.Parent.Save
extractTable Ssilka, book1, iLoop
Set r = .FindNext(r) '<--| the FindNext() method is still called on the same range as in the preceding .Find() statement
Loop While Not r Is Nothing And r.Address <> firstAddress And iLoop < 19 '<--| exit loop if either you hit the first link or completed three loops
End If
End With
book1.Save
book1.Close
Exit Sub
End Sub
Function extractTable(Ssilka As String, book1 As Workbook, iLoop As Long)
Dim oDom As Object, oTable As Object, oRow As Object
Dim iRows As Integer, iCols As Integer
Dim x As Integer, y As Integer
Dim data()
Dim oHttp As Object
Dim oRegEx As Object
Dim sResponse As String
Dim oRange As Range
' get page
Set oHttp = CreateObject("MSXML2.XMLHTTP")
oHttp.Open "GET", Ssilka, False
oHttp.Send
' cleanup response
sResponse = StrConv(oHttp.responseBody, vbUnicode)
Set oHttp = Nothing
sResponse = Mid$(sResponse, InStr(1, sResponse, "<!DOCTYPE "))
Set oRegEx = CreateObject("vbscript.regexp")
With oRegEx
.MultiLine = True
.Global = True
.IgnoreCase = False
.Pattern = "<(script|SCRIPT)[\w\W]+?</\1>"
sResponse = .Replace(sResponse, "")
End With
Set oRegEx = Nothing
' create Document from response
Set oDom = CreateObject("htmlFile")
oDom.Write sResponse
DoEvents
' table with results, indexes starts with zero
Set oTable = oDom.getelementsbytagname("table")(3)
DoEvents
iRows = oTable.Rows.Length
iCols = oTable.Rows(1).Cells.Length
' first row and first column contain no intresting data
ReDim data(1 To iRows - 1, 1 To iCols - 1)
' fill in data array
For x = 1 To iRows - 1
Set oRow = oTable.Rows(x)
For y = 1 To iCols - 1
If oRow.Cells(y).Children.Length > 0 Then
data(x, y) = oRow.Cells(y).getelementsbytagname("a")(0).getattribute("href")
'.Replace(data(x, y), "about:", "http://allscores.ru/soccer/")
End If
Next y
Next x
Set oRow = Nothing
Set oTable = Nothing
Set oDom = Nothing
' put data array on worksheet
Set oRange = book1.ActiveSheet.Cells(34, iLoop * 25).Resize(iRows - 1, iCols - 1)
oRange.NumberFormat = "@"
oRange.Value = data
oRange.Replace What:="about:", Replacement:="http://allscores.ru/soccer/"
Set oRange = Nothing
'Selection.Replace What:=".", Replacement:=",", LookAt:=xlPart, _
SearchOrder:=xlByRows, MatchCase:=False, SearchFormat:=False, _
ReplaceFormat:=False, MatchByte:=False
'<DEBUG>
' For x = LBound(data) To UBound(data)
' Debug.Print x & ":[ ";
' For y = LBound(data, 2) To UBound(data, 2)
' Debug.Print y & ":[" & data(x, y) & "] ";
' Next y
' Debug.Print "]"
' Next x
'</DEBUG>
End Function
答案 0 :(得分:1)
正如@ YowE3K的评论中提到的,如果r is Nothing
,VBA引擎将继续评估IF语句,并在r.Address
上失败。
其他语言的行为有所不同,一旦发现错误情况就会逃避检查,但VBA不会这样做 - 这称为短路评估 - {{3} }
这是一种解决方法:
Option Explicit
Public Sub TestMe()
Dim iloop As Long
Dim r As Range
Dim firstAddress As String
Do While True
If r Is Nothing Then Exit Do
If r.Address = firstAddress Then Exit Do
If iloop < 10 Then Exit Do
'Do the action
Loop
End Sub