How to handle errors efficiently to prevent misleading results? - vba

Question

I've written some code in vba to find certain identities against some names in some websites. The code is working well if everything in it's right order, i meant if the link is valid, the name matches with a tags and finally the regex can find the identity. If any of the three or all of the three are bad searches then the script throws error. I've already specified the position where error occurs in my below script.
All i expect from you experts to provide me with any solution as to how i can handle the errors and let my script continue until all the links are exhausted.
As I do not have much knowledge on VBA so i tried with On error resume next to skip the errors. However, it turns out to be a clear mess when i take a look at the results. I'm pasting a rough example what i get when i use On error resume next.
Sub Identity_Finder()
Dim http As New XMLHTTP60, html As New HTMLDocument
Dim post As Object, link As Variant, refined_links As String
Dim rxp As New RegExp, identity As Object
For Each link In [{"http://spltech.in/","http://www.unifrostindia.com/","http://www.unitfrostindia.com/","http://www.greenplanet.in/"}]
With http
.Open "GET", link, False
.send '''throws here the first error if the link is invalid
html.body.innerHTML = .responseText
End With
For Each post In html.getElementsByTagName("a")
If InStr(post.innerText, "certain_name") > 0 Then refined_links = post.href: Exit For
Next post
With http
.Open "GET", refined_links, False
.send ''throws another error here if no such link is found
End With
With rxp
.Pattern = "some_regex"
.Global = True
Set identity = .Execute(http.responseText)
End With
r = r + 1: Cells(r, 1) = link
Cells(r, 2) = identity(0) ''''throws another error here if no such identity is noticed
Next link
End Sub
Upon using On error resume next What i get:
John executive
Mac lawyer
lulu lawyer
Robin lawyer
Cathy student
Expected output:
John executive
Mac lawyer
lulu
Robin
Cathy student
The empty fields (when they are not found) are getting filled in with the previous values when i use On error resume next. How can I get around this misleading result? Thanks in advance.

Answer 1

The most efficient way to error trap in VBA is to
1) actually test the inputs / results before running them either through custom-made functions or built-in coding concepts or a mix of both.
2) Use VBA built-in error-handling if absolutely needed
Example 1
For example. You can wrap this statement with a custom function to test if a URL is valid or not.
With http
.Open "GET", link, False
.send '''throws here the first error if the link is invalid
html.body.innerHTML = .responseText
End With
If ValidURL Then
With http
.Open "GET", link, False
.send
html.body.innerHTML = .responseText
End With
End If
Where ValidURL is a function defined as:
Function ValidURL(URL as String) as Boolean
Dim result as Boolean
'I don't know how you would specify a valid link in your specific case
'but that code goes here
'a dummy example follows
result = Left(URL,7) = "http://"
ValidURL = result 'True or False
End Function
Example 2
I assume in this statement:
With http
.Open "GET", refined_links, False
.send ''throws another error here if no such link is found
End With
there is a specific error number (code) that is produced when no such link is found. Discover that number and use this code to bypass.
With http
.Open "GET", refined_links, False
On Error Resume Next
.Send
On Error GoTo 0
End With
If err.Number <> 9999 'replace with correct number
'continue with regex test
End If
PUTTING IT ALL TOGETHER
Finally putting that all together you can build like so, with minimal use of On Error Resume Next and no GoTo statements.
For Each link In [{"http://spltech.in/","http://www.unifrostindia.com/","http://www.unitfrostindia.com/","http://www.greenplanet.in/"}]
If ValidURL(link) Then
With http
.Open "GET", link, False
.send
html.body.innerHTML = .responseText
End With
For Each post In html.getElementsByTagName("a")
If InStr(post.innerText, "certain_name") > 0 Then refined_links = post.href: Exit For
Next post
With http
.Open "GET", refined_links, False
On Error Resume Next
.Send
On Error GoTo 0
End With
If err.Number <> 9999 'replace with correct number
With rxp
.Pattern = "some_regex"
.Global = True
Set identity = .Execute(http.responseText)
End With
'i will leave it to you on how to account for no pattern match
r = r + 1: Cells(r, 1) = link
Cells(r, 2) = identity(0) ''''throws another error here if no such identity is noticed
End If
End If
Next link