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Cannot Dechiper Encoded String
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 |
Cannot Dechiper Encoded String
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$(i + k)) Next Next MsgBox Subject ' Auction End Sub Regards, Peter T "Dudely" wrote in message ... I pull a text string off a web page, that contains some random text. From time to time, the string will contain a mixture of ASCII and ISO encoded characters. Sample below. I'm unable to get any function to recognize the string. I'm using VBA 6.0 with Excel 2000 Example code: Subject= "Auction" Subject = Replace(Subject, "B", "A") I've also tried every other character that should be in that string. The replacement never takes place, the string is unmodified. I've also tried searching for just "&#" but it never finds that either. I've tried using InStr as well but that fails to find a match too. The goal is of course to dechiper the (random) string into it's ASCII equivalent (so that it can be read by a human). Any help? Thank you in advance - Dudely |
Cannot Dechiper Encoded String
Yes, the typo was in the post, not in the original code.
I didn't quite understand what you were doing, so I changed your code a bit to: For i = 48 To 127 sfind = Replace("&#num;", "num", CStr(i)) Subject = Replace(Subject, sfind, Chr$(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$(i + k)) * * * * Next * * Next * * MsgBox Subject ' Auction End Sub Regards, Peter T "Dudely" wrote in message ... I pull a text string off a web page, that contains some random text. From time to time, the string will contain a mixture of ASCII and ISO encoded characters. *Sample below. I'm unable to get any function to recognize the string. I'm using VBA 6.0 with Excel 2000 Example code: Subject= "Auction" Subject = Replace(Subject, "B", "A") I've also tried every other character that should be in that string. The replacement never takes place, the string is unmodified. *I've also tried searching for just "&#" but it never finds that either. I've tried using InStr as well but that fails to find a match too. The goal is of course to dechiper the (random) string into it's ASCII equivalent (so that it can be read by a human). Any help? Thank you in advance - Dudely- Hide quoted text - - Show quoted text - |
Cannot Dechiper Encoded String
The double loop I posted processed char codes 65-90 and 97-122, ie A-Z,a-z.
Even with the limited information and typo you posted it appeared to successfully decode your sample string. If it fails to decode different sample data, it's impossible for anyone to assist without that data, and ideally what the sample should decode as (doesn't look like it's encrypted). Regards, Peter T "Dudely" wrote in message ... Yes, the typo was in the post, not in the original code. I didn't quite understand what you were doing, so I changed your code a bit to: For i = 48 To 127 sfind = Replace("&#num;", "num", CStr(i)) Subject = Replace(Subject, sfind, Chr$(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$(i + k)) Next Next MsgBox Subject ' Auction End Sub Regards, Peter T "Dudely" wrote in message ... I pull a text string off a web page, that contains some random text. From time to time, the string will contain a mixture of ASCII and ISO encoded characters. Sample below. I'm unable to get any function to recognize the string. I'm using VBA 6.0 with Excel 2000 Example code: Subject= "Auction" Subject = Replace(Subject, "B", "A") I've also tried every other character that should be in that string. The replacement never takes place, the string is unmodified. I've also tried searching for just "&#" but it never finds that either. I've tried using InStr as well but that fails to find a match too. The goal is of course to dechiper the (random) string into it's ASCII equivalent (so that it can be read by a human). Any help? Thank you in advance - Dudely- Hide quoted text - - Show quoted text - |
Cannot Dechiper Encoded String
Thanks for your response. Here is a cut & paste of another sample
string retrieved. Bank Seized I've changed your code to do the following: For i = 0 To 255 sfind = Replace("&#num;", "num", CStr(i)) Subject = Replace(Subject, sfind, Chr$(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 - |
Cannot Dechiper Encoded String
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 - |
Cannot Dechiper Encoded String
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Cannot Dechiper Encoded String
On Sun, 16 Nov 2008 09:00:24 -0800 (PST), Dudely wrote:
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 , then the corresponding values would be: ?subject=B ank 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 I suspect something odd about the code you have not posted. Perhaps if you posted all of your code, and not just bits and pieces, it might be helpful. The following seems to work as expected: ===================== Option Explicit Const e1 As String = _ ?subject=Bank Seized" Sub foo() Dim SubjectString As String Dim i As Long Dim theSpot As Long theSpot = InStr(e1, "subject") SubjectString = Mid(e1, theSpot + 8) For i = 32 To 127 SubjectString = Replace(SubjectString, "&#" & i & ";", Chr(i)) Next i Debug.Print SubjectString End Sub ==================== --ron |
Cannot Dechiper Encoded String
On Sun, 16 Nov 2008 09:00:24 -0800 (PST), Dudely wrote:
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 , then the corresponding values would be: ?subject=B ank 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 As I mentioned -- something in your code. The above code, which you did post, is ambiguous. You mention in a previous post that e1 is of type variant. That being the case, e1.value is not valid and should be giving you an error. --ron |
Cannot Dechiper Encoded String
Sorry, I didn't express myself clearly enough.
The value of el, as seen in the watch window in the value column, is what I meant when I typed el.value. I'm actually accessing "el", not "el.value'. I apologize for the confusion. Also, I recently realized that the debug window is showing ?subject=Got%20ize %20" for el, whereas when I look at the web page source, that same line of data that I'm capturing as "el", shows up as:Gov't Seized So that explains why it's not working for me. By the time the datastring is captured by a variable something has already discarded the characters that need translating. I'm getting el this way: Set IeDoc = obIe.document el = IeDoc.getElementsByTagName("HR") el = el.nextSibling.nextSibling Here's where I started to copy the data from the web page source, and that's when I realized - the data I'm trying to translate doesn't exist in the line of data I'm capturing! It had already been removed! No wonder it didn't work! So, from there it was easy to figure out. I needed to grab that data from a different element. And I did. And now it works. Many thanks to both of you for your help, you led me to the actual problem. Couldn't of done it without either of you. On Nov 16, 9:40*am, Ron Rosenfeld wrote: On Sun, 16 Nov 2008 09:00:24 -0800 (PST), Dudely wrote: 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 , then the corresponding values would be: ?subject=B ank 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 I suspect something odd about the code you have not posted. *Perhaps if you posted all of your code, and not just bits and pieces, it might be helpful. The following seems to work as expected: ===================== Option Explicit Const e1 As String = _ ?subject=Bank Seized" Sub foo() Dim SubjectString As String Dim i As Long Dim theSpot As Long * * theSpot = InStr(e1, "subject") * * SubjectString = Mid(e1, theSpot + 8) For i = 32 To 127 * * SubjectString = Replace(SubjectString, "&#" & i & ";", Chr(i)) Next i Debug.Print SubjectString End Sub ==================== --ron |
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