460,028 Members | 1,066 Online
Need help? Post your question and get tips & solutions from a community of 460,028 IT Pros & Developers. It's quick & easy.

# Help on Regular Expression

 P: n/a I am new in Regular Expression. Could someone please help me in following expression? 1. the string cannot be empty 2. the string can only contains AlphaNumeric characters. No space or any special characters are allowed 3. space characters at the end of string is ok 4. the string cannot contains only numeric characters, in other word, the string must contains a least one alpha character Thanks for the help Nov 21 '05 #1
5 Replies

 P: n/a John: The following code will check for a string that is 1-20 characters long. It does not specifically test Condition 3 but you could trim the string and then check it. Private Function CheckStringRules(ByVal psString) As _Outcome Dim lsPattern As String Dim mc As MatchCollection Dim m As Match 'Check for allowed character set; the length has to be 1-20 lsPattern = "^([a-zA-Z0-9|]{1,20})\$" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of number lsPattern = "[0-9]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of upper-alpha character lsPattern = "[A-Z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of lower-alpha character lsPattern = "[a-z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If Return _Outcome.Successful End Function "John" wrote: I am new in Regular Expression. Could someone please help me in following expression? 1. the string cannot be empty 2. the string can only contains AlphaNumeric characters. No space or any special characters are allowed 3. space characters at the end of string is ok 4. the string cannot contains only numeric characters, in other word, the string must contains a least one alpha character Thanks for the help Nov 21 '05 #2

 P: n/a Thanks vvenk for your reply. What I want to do is to add a RegularExpressionValidator control in the form and specify a RegularExpression for the Validator control and validate against a textbox in client side. Is that possible to combine all the requirement into one Regular Expression, so that I don't need to go to code-behind to do the validation? Thanks again. "vvenk" wrote in message news:45**********************************@microsof t.com... John: The following code will check for a string that is 1-20 characters long. It does not specifically test Condition 3 but you could trim the string and then check it. Private Function CheckStringRules(ByVal psString) As _Outcome Dim lsPattern As String Dim mc As MatchCollection Dim m As Match 'Check for allowed character set; the length has to be 1-20 lsPattern = "^([a-zA-Z0-9|]{1,20})\$" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of number lsPattern = "[0-9]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of upper-alpha character lsPattern = "[A-Z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of lower-alpha character lsPattern = "[a-z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If Return _Outcome.Successful End Function "John" wrote: I am new in Regular Expression. Could someone please help me in following expression? 1. the string cannot be empty 2. the string can only contains AlphaNumeric characters. No space or any special characters are allowed 3. space characters at the end of string is ok 4. the string cannot contains only numeric characters, in other word, the string must contains a least one alpha character Thanks for the help Nov 21 '05 #3

 P: n/a Aren't Regular Expressions powerful ... (?i)^([a-z]|\d)+((?<=\d)[a-z]|(?<=[a-z]))[a-z0-9]* case insensitive start of string match one or more letters OR one or more digits continue matching if previous is one digit followed by one letter OR one letter match zero or more letters and digits --Robby "John" wrote in message news:um*************@TK2MSFTNGP11.phx.gbl... Thanks vvenk for your reply. What I want to do is to add a RegularExpressionValidator control in the form and specify a RegularExpression for the Validator control and validate against a textbox in client side. Is that possible to combine all the requirement into one Regular Expression, so that I don't need to go to code-behind to do the validation? Thanks again. "vvenk" wrote in message news:45**********************************@microsof t.com... John: The following code will check for a string that is 1-20 characters long. It does not specifically test Condition 3 but you could trim the string and then check it. Private Function CheckStringRules(ByVal psString) As _Outcome Dim lsPattern As String Dim mc As MatchCollection Dim m As Match 'Check for allowed character set; the length has to be 1-20 lsPattern = "^([a-zA-Z0-9|]{1,20})\$" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of number lsPattern = "[0-9]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of upper-alpha character lsPattern = "[A-Z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of lower-alpha character lsPattern = "[a-z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If Return _Outcome.Successful End Function "John" wrote: > I am new in Regular Expression. Could someone please help me in following > expression? > 1. the string cannot be empty > 2. the string can only contains AlphaNumeric characters. No space or > any > special characters are allowed > 3. space characters at the end of string is ok > 4. the string cannot contains only numeric characters, in other word, the > string must contains a least one alpha character > > Thanks for the help > > > Nov 21 '05 #4

 P: n/a opps ... Forgot to check remaining characters are spaces. (?i)^([a-z]|\d)+((?<=\d)[a-z]|(?<=[a-z]))[a-z0-9]* *\$ That is a space between the two * " *" matches zero or more spaces \$ matches end of string --Robby "Robby" wrote in message news:u%****************@TK2MSFTNGP09.phx.gbl... Aren't Regular Expressions powerful ... (?i)^([a-z]|\d)+((?<=\d)[a-z]|(?<=[a-z]))[a-z0-9]* case insensitive start of string match one or more letters OR one or more digits continue matching if previous is one digit followed by one letter OR one letter match zero or more letters and digits --Robby "John" wrote in message news:um*************@TK2MSFTNGP11.phx.gbl... Thanks vvenk for your reply. What I want to do is to add a RegularExpressionValidator control in the form and specify a RegularExpression for the Validator control and validate against a textbox in client side. Is that possible to combine all the requirement into one Regular Expression, so that I don't need to go to code-behind to do the validation? Thanks again. "vvenk" wrote in message news:45**********************************@microsof t.com... John: The following code will check for a string that is 1-20 characters long. It does not specifically test Condition 3 but you could trim the string and then check it. Private Function CheckStringRules(ByVal psString) As _Outcome Dim lsPattern As String Dim mc As MatchCollection Dim m As Match 'Check for allowed character set; the length has to be 1-20 lsPattern = "^([a-zA-Z0-9|]{1,20})\$" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of number lsPattern = "[0-9]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of upper-alpha character lsPattern = "[A-Z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of lower-alpha character lsPattern = "[a-z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If Return _Outcome.Successful End Function "John" wrote: > I am new in Regular Expression. Could someone please help me in following > expression? > 1. the string cannot be empty > 2. the string can only contains AlphaNumeric characters. No space or > any > special characters are allowed > 3. space characters at the end of string is ok > 4. the string cannot contains only numeric characters, in other word, the > string must contains a least one alpha character > > Thanks for the help > > > Nov 21 '05 #5

 P: n/a Robby, I am really appreciated for your help! "Robby" wrote in message news:OB**************@TK2MSFTNGP09.phx.gbl... opps ... Forgot to check remaining characters are spaces. (?i)^([a-z]|\d)+((?<=\d)[a-z]|(?<=[a-z]))[a-z0-9]* *\$ That is a space between the two * " *" matches zero or more spaces \$ matches end of string --Robby "Robby" wrote in message news:u%****************@TK2MSFTNGP09.phx.gbl... Aren't Regular Expressions powerful ... (?i)^([a-z]|\d)+((?<=\d)[a-z]|(?<=[a-z]))[a-z0-9]* case insensitive start of string match one or more letters OR one or more digits continue matching if previous is one digit followed by one letter OR one letter match zero or more letters and digits --Robby "John" wrote in message news:um*************@TK2MSFTNGP11.phx.gbl... Thanks vvenk for your reply. What I want to do is to add a RegularExpressionValidator control in the form and specify a RegularExpression for the Validator control and validate against a textbox in client side. Is that possible to combine all the requirement into one Regular Expression, so that I don't need to go to code-behind to do the validation? Thanks again. "vvenk" wrote in message news:45**********************************@microsof t.com... John: The following code will check for a string that is 1-20 characters long. It does not specifically test Condition 3 but you could trim the string and then check it. Private Function CheckStringRules(ByVal psString) As _Outcome Dim lsPattern As String Dim mc As MatchCollection Dim m As Match 'Check for allowed character set; the length has to be 1-20 lsPattern = "^([a-zA-Z0-9|]{1,20})\$" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of number lsPattern = "[0-9]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of upper-alpha character lsPattern = "[A-Z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If 'Check for atleast one occurence of lower-alpha character lsPattern = "[a-z]" mc = Regex.Matches(psString, lsPattern) If mc.Count = 0 Then Return _Outcome.Failure End If Return _Outcome.Successful End Function "John" wrote: > I am new in Regular Expression. Could someone please help me in following > expression? > 1. the string cannot be empty > 2. the string can only contains AlphaNumeric characters. No space or > any > special characters are allowed > 3. space characters at the end of string is ok > 4. the string cannot contains only numeric characters, in other word, the > string must contains a least one alpha character > > Thanks for the help > > > Nov 21 '05 #6

### This discussion thread is closed

Replies have been disabled for this discussion.