Escaping problem using Regular Expression

[Henry]

I have this simple code,

string escaped = Regex.Escape( @"`~!@#$%^&*()_=+[]{}\|;:',<.>/?" + "\"" );
string input = @"a&+[b`${c}a'?sd:r]" + "\"" + @"@(-d)\e";
Regex re = new Regex( string.Format(@"([{0}]+)", escaped),
RegexOptions.CultureInvariant );
string s = re.Replace( input, "" );

It doesn't seem to work, regular expression return without filter out any
character
However if I remove or change the position of "]" to be the first character,
it works.

string escaped = Regex.Escape( @"]`~!@#$%^&*()_=+[{}\|;:',<.>/?" + "\"" );

I totally do not understand how this regular expression escaping works. What
am I doing wrong here?

Thanks
Henry

[Nick Malik]

Hi Henry,
Regular expressions are made up of many of the characters that you placed
into your string. Including the [ and ] characters. The pattern that you
specify has an invalid part [] which is a pattern group where any member can
be matched, but it has no members. that makes the pattern invalid.

Moving the ] to the front of the string means that the regex parser will not
see the bad pattern. However to be truly correct, you still need to quote
any of the characters that can be interpreted as part of a pattern.

@"`~\!@#\$\%\^\&\*\(\)_=+\[\]\{\}\\|;:',<.>/?" + "\""

Hope this helps,
--- Nick

"Henry" <ig******@hotmail.com> wrote in message
news:%2****************@tk2msftngp13.phx.gbl...

I have this simple code,

string escaped = Regex.Escape( @"`~!@#$%^&*()_=+[]{}\|;:',<.>/?" + "\"" );
string input = @"a&+[b`${c}a'?sd:r]" + "\"" + @"@(-d)\e";
Regex re = new Regex( string.Format(@"([{0}]+)", escaped),
RegexOptions.CultureInvariant );
string s = re.Replace( input, "" );

It doesn't seem to work, regular expression return without filter out any
character
However if I remove or change the position of "]" to be the first character, it works.

string escaped = Regex.Escape( @"]`~!@#$%^&*()_=+[{}\|;:',<.>/?" + "\"" );

I totally do not understand how this regular expression escaping works. What am I doing wrong here?

Thanks
Henry

[Martin Marinov]

Hi Henry,

in addition of Nick's answer, there is a little modification

@"`~\!@#\$\%\^\&\*\(\)_=\+\[\]\{\}\\\|;:',<\.>/\?" + "\""

Regards
Martin
"Nick Malik" <ni*******@hotmail.nospam.com> wrote in message
news:OcKFc.21331$Oq2.9279@attbi_s52...

Hi Henry,
Regular expressions are made up of many of the characters that you placed
into your string. Including the [ and ] characters. The pattern that you
specify has an invalid part [] which is a pattern group where any member can be matched, but it has no members. that makes the pattern invalid.

Moving the ] to the front of the string means that the regex parser will not see the bad pattern. However to be truly correct, you still need to quote
any of the characters that can be interpreted as part of a pattern.

@"`~\!@#\$\%\^\&\*\(\)_=+\[\]\{\}\\|;:',<.>/?" + "\""

Hope this helps,
--- Nick

"Henry" <ig******@hotmail.com> wrote in message
news:%2****************@tk2msftngp13.phx.gbl...

I have this simple code,

string escaped = Regex.Escape( @"`~!@#$%^&*()_=+[]{}\|;:',<.>/?" + "\"" ); string input = @"a&+[b`${c}a'?sd:r]" + "\"" + @"@(-d)\e";
Regex re = new Regex( string.Format(@"([{0}]+)", escaped),
RegexOptions.CultureInvariant );
string s = re.Replace( input, "" );

It doesn't seem to work, regular expression return without filter out any character
However if I remove or change the position of "]" to be the first

character,

it works.

string escaped = Regex.Escape( @"]`~!@#$%^&*()_=+[{}\|;:',<.>/?" + "\"" );
I totally do not understand how this regular expression escaping works.

What

am I doing wrong here?

Thanks
Henry