Ron has more than covered anything I can add in response, in particular el
vs el.value
FWIW you should be able to pass the original string to the de-crypt routine
and return a sensible result (ie before parsing with InStr)
Regards,
Peter T
"Dudely" <(E-Mail Removed)> wrote in message
news:8f2c3fcf-7bdf-47b5-b5fe-(E-Mail Removed)...
The string value of el is composed of letters & numbers that make up
an email address, +1. It comes in the form of: *@*.* and is composed
of a random number of characters followed by ?subject= followed by the
actual subject string. The value of theSpot is equal to the number of
characters in the email address. If in fact the email address were
(E-Mail Removed), then the corresponding values would be:
el.value=mailto

(E-Mail Removed)?subject=Bank
Seized
theSpot=the number of characters between the start of the value and
the word "subject" and is obtained by the expression: theSpot = InStr
(el, "subject")
The subject string=Bank Seized
On Nov 16, 2:11 am, "Peter T" <peter_t@discussions> wrote:
> What is the Subject string after doing this
>
> Subject = Mid(el, theSpot + 8)
>
> What is the string value of 'el' and the numeric value of 'theSpot'
>
> Regards,
> Peter T
>
> "Dudely" <ab3...@gmail.com> wrote in message
>
> news:1866ec06-e510-40d3-ab15-(E-Mail Removed)...
> Thanks for your response. Here is a cut & paste of another sample
> string retrieved.
>
> Bank Seized
>
> I've changed your code to do the following:
>
> For i = 0 To 255
> sfind = Replace("&#num;", "num", CStr(i))
> Subject = Replace(Subject, sfind, Chr$(i))
> Next
>
> When the string is copied to a static variable; i.e.
> str="Bank Seized" for the purposes of testing, then
> there is no problem.
> However, when I pull it directly off the web page by accessing the
> DOM, and then parse it like so: Subject = Mid(el, theSpot + 8), the
> above code fails to make the replacement. Note that "el" is of type
> variant. Perhaps this has something to do with it? Subject is of
> type String.
>
> Thank you again for your help.
>
> On Nov 12, 1:50 am, "Peter T" <peter_t@discussions> wrote:
>
>
>
> > The double loop I posted processed char codes 65-90 and 97-122, ie
> > A-Z,a-z.
>
> > Even with the limited information and typo you posted it appeared to
> > successfully decode your sample string.
>
> > If it fails to decode different sample data, it's impossible for anyone
> > to
> > assist without that data, and ideally what the sample should decode as
> > (doesn't look like it's encrypted).
>
> > Regards,
> > Peter T
>
> > "Dudely" <ab3...@gmail.com> wrote in message
>
> >news:8006a934-6ef7-46d3-8912-(E-Mail Removed)...
> > Yes, the typo was in the post, not in the original code.
>
> > I didn't quite understand what you were doing, so I changed your code
> > a bit to:
>
> > For i = 48 To 127
> > sfind = Replace("&#num;", "num", CStr(i))
> > Subject = Replace(Subject, sfind, Chr$(i))
> > Next
> > which I believe captures the essence of what you were doing. The
> > target string could be almost anything and must be able to handle any
> > character. For the moment, I'm assuming the ASCII character set to be
> > sufficient, but in truth there's nothing to prevent someone from using
> > alphanum characters outside that range as well.
>
> > In any event, while the sample string works, live data continues to
> > fail.
>
> > What I'd prefer to do, is setup a "capture", because it walks through
> > a LOT of data before it gets to one of these encoded text strings and
> > I'd rather not have to check every single iteration to see if it has
> > one of the target strings.
>
> > What I mean is that instead of trying to fix it first, I'd prefer to
> > find and capture the text. Then later a similar method can be used to
> > replace the target string.
>
> > So instead of the above, it might say something like:
>
> > if inStr(1, Subject, "&#") then
> > msgBox "Found " + Subject
> > endif
>
> > Something like that. Note, that I tried something like that and it
> > also failed.
>
> > Thank you most kindly for your help.
>
> > - Dudely
>
> > On Nov 11, 12:18 am, "Peter T" <peter_t@discussions> wrote:
>
> > > Looks like you have a typo, 66 vs 65
>
> > > Try this too -
>
> > > Sub test()
> > > Dim i As Long, j As Long, k As Long
> > > Dim Subject As String
>
> > > Subject = "Auction"
>
> > > For j = 0 To 1
> > > k = 32 * j
> > > For i = 65 To 65 + 26
> > > sfind = Replace("&#num;", "num", CStr(i + k))
> > > Subject = Replace(Subject, sfind, Chr$(i + k))
> > > Next
> > > Next
> > > MsgBox Subject ' Auction
> > > End Sub
>
> > > Regards,
> > > Peter T
>
> > > "Dudely" <ab3...@gmail.com> wrote in message
>
> > >news:f1da081c-c0c3-41a5-8204-(E-Mail Removed)...
>
> > > >I pull a text string off a web page, that contains some random text.
> > > > From time to time, the string will contain a mixture of ASCII and
> > > > ISO
> > > > encoded characters. Sample below.
>
> > > > I'm unable to get any function to recognize the string.
>
> > > > I'm using VBA 6.0 with Excel 2000
>
> > > > Example code:
>
> > > > Subject= "Auction"
>
> > > > Subject = Replace(Subject, "B", "A")
>
> > > > I've also tried every other character that should be in that string.
>
> > > > The replacement never takes place, the string is unmodified. I've
> > > > also tried searching for just "&#" but it never finds that either.
> > > > I've tried using InStr as well but that fails to find a match too.
> > > > The goal is of course to dechiper the (random) string into it's
> > > > ASCII
> > > > equivalent (so that it can be read by a human).
>
> > > > Any help?
>
> > > > Thank you in advance
>
> > > > - Dudely- Hide quoted text -
>
> > > - Show quoted text -- Hide quoted text -
>
> > - Show quoted text -- Hide quoted text -
>
> - Show quoted text -