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

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" wrote in message
...
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

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" wrote in message

...



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 -

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" wrote in message
...
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" wrote in message

...



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 -

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" wrote in message

...
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" wrote in message

...

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 -

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" wrote in message
...
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" wrote in message

...
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" wrote in message

...

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 -