The title says it:
I have an excel Sheet with an column full of hyperlinks. Now I want that an VBA Script checks which hyperlinks are dead or work and makes an entry into the next columns either with the text 404 Error or active.
Hopefully someone can help me because I am not really good at VB.
EDIT:
I found @ http://www.utteraccess.com/forums/printthread.php?Cat=&Board=84&main=1037294&type=thread
A solution which is made for word but the Problem is that I need this solution for Excel. Can someone translate this to Excel solution?
Private Sub testHyperlinks()
Dim thisHyperlink As Hyperlink
For Each thisHyperlink In ActiveDocument.Hyperlinks
If thisHyperlink.Address <> "" And Left(thisHyperlink.Address, 6) <> "mailto" Then
If Not IsURLGood(thisHyperlink.Address) Then
Debug.Print thisHyperlink.Address
End If
End If
Next
End Sub
Private Function IsURLGood(url As String) As Boolean
' Test the URL to see if it is good
Dim request As New WinHttpRequest
On Error GoTo IsURLGoodError
request.Open "GET", url
request.Send
If request.Status = 200 Then
IsURLGood = True
Else
IsURLGood = False
End If
Exit Function
IsURLGoodError:
IsURLGood = False
End Function
First add a reference to Microsoft XML V3 (or above), using Tools->References. Then paste this code:
Option Explicit
Sub CheckHyperlinks()
Dim oColumn As Range
Set oColumn = GetColumn() ' replace this with code to get the relevant column
Dim oCell As Range
For Each oCell In oColumn.Cells
If oCell.Hyperlinks.Count > 0 Then
Dim oHyperlink As Hyperlink
Set oHyperlink = oCell.Hyperlinks(1) ' I assume only 1 hyperlink per cell
Dim strResult As String
strResult = GetResult(oHyperlink.Address)
oCell.Offset(0, 1).Value = strResult
End If
Next oCell
End Sub
Private Function GetResult(ByVal strUrl As String) As String
On Error Goto ErrorHandler
Dim oHttp As New MSXML2.XMLHTTP30
oHttp.Open "HEAD", strUrl, False
oHttp.send
GetResult = oHttp.Status & " " & oHttp.statusText
Exit Function
ErrorHandler:
GetResult = "Error: " & Err.Description
End Function
Private Function GetColumn() As Range
Set GetColumn = ActiveWorkbook.Worksheets(1).Range("A:A")
End Function
Gary's code is perfect, but I would rather use a public function in a module and use it in a cell as function. The advantage is that you can use it in a cell of your choice or anyother more complex function.
In the code below I have adjusted Gary's code to return a boolean and you can then use this output in an =IF(CHECKHYPERLINK(A1);"OK";"FAILED"). Alternatively you could return an Integer and return the status itself (eg.: =IF(CHECKHYPERLINK(A1)=200;"OK";"FAILED"))
A1: http://www.whatever.com
A2: =IF(CHECKHYPERLINK(A1);"OK";"FAILED")
To use this code please follow Gary's instructions and additionally add a module to the workbook (right click on the VBAProject --> Insert --> Module) and paste the code into the module.
Option Explicit
Public Function CheckHyperlink(ByVal strUrl As String) As Boolean
Dim oHttp As New MSXML2.XMLHTTP30
On Error GoTo ErrorHandler
oHttp.Open "HEAD", strUrl, False
oHttp.send
If Not oHttp.Status = 200 Then CheckHyperlink = False Else CheckHyperlink = True
Exit Function
ErrorHandler:
CheckHyperlink = False
End Function
Please also be aware that, if the page is down, the timeout can be long.
