Escaping problem using Regular Expression

H

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
 
N

Niki Estner

Henry said:
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, "" );
Click to expand...


Escaping rules within '[' ']''s are different: AFAIK only \], \- and control
characters (\r, \n...) have to be escaped in such a block.
It doesn't seem to work, regular expression return without filter out any
character
Click to expand...


Try matching for new Regex(@"[`~!@#$%^&*()_=+[\]{}\|;:' said:
However if I remove or change the position of "]" to be the first character,
it works.
Click to expand...

Yep, in "normal" regex context "]" is not a metacharacter, so you don't have
to escape it. That is, "Escape(...)" won't escape it either. That should
explain it.
I totally do not understand how this regular expression escaping works. What
am I doing wrong here?
Click to expand...

Use "Escape(...)" if you match for an exact string. You can't put the
escaped sequence in a [..] block. That's it.

Niki
 
H

Henry

Thanks for answering.
Now I got it working. I modified my code as follows:

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


But I still don't really get what's the problem cause in my other code it
doesn't work as what my understanding.

string QuantifierRegex = @"(?=^[\w$#]{7,11}$)";
//string password = "`~!@#$%^&*";
string password = "abc4ddf";

Regex re1 = new Regex( QuantifierRegex );
bool isMatch1 = re1.IsMatch( password );

Regex re2 = new Regex( Regex.Escape(QuantifierRegex) );
bool isMatch2 = re2.IsMatch( password );

bool isMatch3 = Regex.IsMatch( password, QuantifierRegex,
RegexOptions.CultureInvariant );
bool isMatch4 = Regex.IsMatch( password, Regex.Escape(QuantifierRegex),
RegexOptions.CultureInvariant );

From the sample above,
isMatch1 returns true
isMatch2 returns false
isMatch3 returns true
isMatch4 returns false

If I change the password variable to "`~!@#$%^&*"
isMatch1 returns false
isMatch2 returns false
isMatch3 returns false
isMatch4 returns false

Any idea?
Thanks
Henry

Niki Estner said:
Henry said:
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, "" );
Click to expand...


Escaping rules within '[' ']''s are different: AFAIK only \], \- and control
characters (\r, \n...) have to be escaped in such a block.
It doesn't seem to work, regular expression return without filter out any
character
Click to expand...


Try matching for new Regex(@"[`~!@#$%^&*()_=+[\]{}\|;:' said:
However if I remove or change the position of "]" to be the first character,
it works.
Click to expand...

Yep, in "normal" regex context "]" is not a metacharacter, so you don't have
to escape it. That is, "Escape(...)" won't escape it either. That should
explain it.
I totally do not understand how this regular expression escaping works. What
am I doing wrong here?
Click to expand...

Use "Escape(...)" if you match for an exact string. You can't put the
escaped sequence in a [..] block. That's it.

Niki
Click to expand...
 
N

Niki Estner

Henry said:
...
string QuantifierRegex = @"(?=^[\w$#]{7,11}$)";
//string password = "`~!@#$%^&*";
string password = "abc4ddf";

Regex re1 = new Regex( QuantifierRegex );
bool isMatch1 = re1.IsMatch( password );
Click to expand...

re1 matches a string that consist of 7-11 characters, which may be
alphanumeric (\w), $ or #.
Don't know why you need the paranthesis.
Regex re2 = new Regex( Regex.Escape(QuantifierRegex) );
Click to expand...

re2 matches for literally "(?=^[\w$#]{7,11}$)"
bool isMatch2 = re2.IsMatch( password );

bool isMatch3 = Regex.IsMatch( password, QuantifierRegex,
RegexOptions.CultureInvariant );
bool isMatch4 = Regex.IsMatch( password, Regex.Escape(QuantifierRegex),
RegexOptions.CultureInvariant );

From the sample above,
isMatch1 returns true
isMatch2 returns false
isMatch3 returns true
isMatch4 returns false

If I change the password variable to "`~!@#$%^&*"
isMatch1 returns false
isMatch2 returns false
isMatch3 returns false
isMatch4 returns false

Any idea?
Click to expand...

Yes.
Looks correct.
What did you think it should do?

Niki
 
H

Henry H

Thanks for answering.
Now I got it working. I modified my code as follows:

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


But I still don't really get what's the problem cause in my other code
it doesn't work as what my understanding.

string QuantifierRegex = @"(?=^[\w$#]{7,11}$)";
//string password = "`~!@#$%^&*";
string password = "abc4ddf";

Regex re1 = new Regex( QuantifierRegex );
bool isMatch1 = re1.IsMatch( password );

Regex re2 = new Regex( Regex.Escape(QuantifierRegex) );
bool isMatch2 = re2.IsMatch( password );

bool isMatch3 = Regex.IsMatch( password, QuantifierRegex,
RegexOptions.CultureInvariant );
bool isMatch4 = Regex.IsMatch( password, Regex.Escape(QuantifierRegex),
RegexOptions.CultureInvariant );

From the sample above,
isMatch1 returns true
isMatch2 returns false
isMatch3 returns true
isMatch4 returns false

If I change the password variable to "`~!@#$%^&*"
isMatch1 returns false
isMatch2 returns false
isMatch3 returns false
isMatch4 returns false

Any idea?
Thanks
Henry

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H

Henry

Well, I totally forgot the (\w), $ or #. expression due to the 20 hours I'm
working on the project in a row.

Thanks so much

Niki Estner said:
Henry said:
...
string QuantifierRegex = @"(?=^[\w$#]{7,11}$)";
//string password = "`~!@#$%^&*";
string password = "abc4ddf";

Regex re1 = new Regex( QuantifierRegex );
bool isMatch1 = re1.IsMatch( password );
Click to expand...

re1 matches a string that consist of 7-11 characters, which may be
alphanumeric (\w), $ or #.
Don't know why you need the paranthesis.
Regex re2 = new Regex( Regex.Escape(QuantifierRegex) );
Click to expand...

re2 matches for literally "(?=^[\w$#]{7,11}$)"
bool isMatch2 = re2.IsMatch( password );

bool isMatch3 = Regex.IsMatch( password, QuantifierRegex,
RegexOptions.CultureInvariant );
bool isMatch4 = Regex.IsMatch( password, Regex.Escape(QuantifierRegex),
RegexOptions.CultureInvariant );

From the sample above,
isMatch1 returns true
isMatch2 returns false
isMatch3 returns true
isMatch4 returns false

If I change the password variable to "`~!@#$%^&*"
isMatch1 returns false
isMatch2 returns false
isMatch3 returns false
isMatch4 returns false

Any idea?
Click to expand...

Yes.
Looks correct.
What did you think it should do?

Niki
Click to expand...
 

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