• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 137
  • Last Modified:

parse the next line in outlook

If

If intLocLabel > 0 Then
       strText = Left(Trim(Mid(strSource, intLocLabel + 14)), 9)
       intLenLabel = Len(strText)

parses a 9 digit number just after my key phrase, how would I parse all the information on the next line?
0
bpfsr
Asked:
bpfsr
  • 2
1 Solution
 
Naveen SwamyYash Infinite Solutions Private LimitedCommented:
not clear with your question???????
0
 
Mark_FreeSoftwareCommented:
i think we need to know some more about the formatting....


however to speed this up, you can place $ signs right to Left,Trim and Mid like this:

strText = Left$(Trim$(Mid$(strSource, intLocLabel + 14)), 9)

that is about 20% faster
0
 
bpfsrAuthor Commented:
Sorry - what I'm saying is I use the code above to parse a particular number from the same line as the LocLabel. In this case it is a reference numbers, so I am searching for the phrase "reference #:" and then parsing the number itself, which always falls immediately after "reference #:" and is always on the same line. However, I want to seperately parse, for insertion somewhere else, the line beneath "reference #:". So if  - strText = Left$(Trim$(Mid$(strSource, intLocLabel + 14)), 9) - gives me the number following "reference #:", how do I get the line beneath it?
0
 
Mark_FreeSoftwareCommented:

you can search for a vbcrlf (linebreak)
0

Featured Post

Concerto's Cloud Advisory Services

Want to avoid the missteps to gaining all the benefits of the cloud? Learn more about the different assessment options from our Cloud Advisory team.

  • 2
Tackle projects and never again get stuck behind a technical roadblock.
Join Now