I've searched for hours and already tried tons of different patterns - there's a simple thing I wan't to achive with regex, but somehow it just won't do as I want:
Possible Strings
String1
This is some text \0"ยง%lfsdrlsrblabla\0\0\0}dfglpdfgl
String2
This is some text
String3
This is some text \0
Desired Match/Result
This is some text
I simply want to match everything - until and except the \0 - resulting in only 1 Match. (everything before the \0)
Important for my case is, that it will match everytime, even when the \0 is not given.
Thanks for your help!
You can try with this pattern:
#"^(?:[^\\]+|\\(?!0))+"
In other words: all characters except backslashes or backslashes not followed by 0
I like
#"^((?!\\0).)*"
Because it's very easy to implement with any arbitrary string. The basic trick is the negative lookahead, which asserts that the string starting at this point doesn't match the
regular expression inside. We follow this with a wildcard to mean "Literally any character not at the start of my string. If your string should change, this is an easy update - just
#"^((?!--STRING--).)*)"
As long as you properly escape that string. Heck, with this pattern, you're merely a regex_escape function from generating any delimiter string.
Bonus: using * instead of + will return a blank string as a valid match when your string starts with your delimiter.
Related
I've been trying to figure this out, but I don't think I understand Regex well enough to get to where I need to.
I have string that resemble these:
filename.txt(1)attribute, 2)attribute(s), more!)
otherfile.txt(abc, def)
Basically, a string that always starts with a filename, then has some text between parentheses. And I'm trying to extract that part which is between the main parentheses, but the text that's there can contain absolutely anything, even some more parentheses (it often does.)
Originally, there was a 'hacky' expression made like this:
/\(([^#]+)\)\g
And it worked, until we ran into a case where the input string contained a # and we were stuck. Obviously...
I can't change the way the strings are generated, it's always a filename, then some parentheses and something of unknown length and content inside.
I'm hoping for a simple Regex expression, since I need this to work in both C# and in Perl -- is such a thing possible? Or does this require something more complex, like its own parsing method?
You can change exception for # symbol in your regex to regex matches any characters and add quantifier that matches from 0 to infinity symbols. And also simplify your regex by deleting group construction:
\(.*\)
Here is the explanation for the regular expression:
Symbol \( matches the character ( literally.
.* matches any character (except for line terminators)
* quantifier matches between zero and unlimited times, as many times
as possible, giving back as needed (greedy)
\) matches the character ) literally.
You can use regex101 to compose and debug your regular expressions.
Regex seems overkill to me in this case. Can be more reliably achieved using string manipulation methods.
int first = str.IndexOf("(");
int last = str.LastIndexOf(")");
if (first != -1 && last != -1)
{
string subString = str.Substring(first + 1, last - first - 1);
}
I've never used Perl, but I'll venture a guess that it has equivalent methods.
I'm currently using the following line of code:
Regex Regex_Alpha = new Regex(#"[a-zA-Z]+('[a-zA-Z])?[a-zA-Z]*");
What I want to do is filter the input of text fields with the condition that input should only be letters and the apostrophe symbol (actually, I still want to add more, but I'm trying to resolve this first).
Right now, it is accepting ALL characters, even numbers.
With my understanding of Regex, I tried to formulate my own expression in the line of:
Regex Regex_Alpha = new Regex(#"^[a-zA-Z'-"+$);
It filters numbers, but doesn't accept the apostrophe symbol. Tried to remove the # sign and filter the apostrophe with the backslash escape character, but still no use.
What should be the best approach to filter the input so that it only accepts letters and apostrophe? (I'll do the rest of the symbols once I understand how this one should work)
As I've commented, your first regular expression is a pretty good shot at "letters, with a single apostrophe not at either end". However, it matchs any string with even a single letter because a regular expression looks for any match in the input, not for whether the entire input matches.
You can fix this by doing what you've done in your second regular expression - just put a ^ at the start and a $ at the end. This means the start and end of the expression have to match the start and end of the input, so it ensures the whole input is only made up of letters and a possible apostrophe.
Regarding your second regular expression, you have a few of problems.
If you want a double-quote in a #"..." string literal, you need to put two double quotes. (I think this might just be a typing mistake in your question, as what you currently have wouldn't even compile.)
You need to close your character class with a ], otherwise the [ and everything inside just get treated as a sequence of characters to match, one after the other.
If you want a hyphen in a character class, it has to go at the start or end, or it gets mistaken for a "between" hyphen (as in A-Z).
The expression #"^[a-zA-Z'""-]+$" should match "any string entirely made of letters, apostrophes, quotes or hyphens".
I created the following regex expression for my C# file. Bascily I want the user's input to only be regular characters (A-Z lower or upper) and numbers. (spaces or symbols ).
[a-zA-Z0-9]
For some reason it only fails when its a symbol on its own. if theres characters mixed with it then the expression passes.
I can show you my code of how I implment it but I think its my expression.
Thanks!
The problem is that it can match anywhere. You need anchors:
^[a-zA-Z0-9]+\z
^ matches the start of a string, and \z matches the end of a string.
(Note: in .NET regex, $ matches the end of a string with an optional newline.)
This is because it will match any character in the string you need the following.
Forces it to match the entire string not just part of it
^[0-9a-zA-Z]*$
That regex will match every single alphanumeric character in the string as separate matches.
If you want to make sure the whole string the user entered only has alphanumeric characters you need to do something like:
^[a-zA-Z0-9]+$
Are you making sure to check the whole string? That is are you using an expression like
^[a-zA-Z0-9]*$
where ^ means the start of the string and $ means the end of the string?
I'm currently facing a (little) blocking issue. I'd like to replace a substring by one another using regular expression. But here is the trick : I suck at regex.
Regex.Replace(contenu, "Request.ServerVariables("*"))",
"ServerVariables('test')");
Basically I'd like to replace whatever is between the " by "test". I tried ".{*}" as a pattern but it doesn't work.
Could you give me some tips, I'd appreciate it!
There are several issues you need to take care of.
You are using special characters in your regex (., parens, quotes) -- you need to escape these with a slash. And you need to escape the slashes with another slash as well because we 're in a C# string literal, unless you prefix the string with # in which case the escaping rules are different.
The expression to match "any number of whatever characters" is .*. In this case, you would want to match any number of non-quote characters, which is [^"]*.
In contrast to (1) above, the replacement string is not a regular expression so you don't want any slashes there.
You need to store the return value of the replace somewhere.
The end result is
var result = Regex.Replace(contenu,
#"Request\.ServerVariables\(""[^""]*""\)",
"Request.ServerVariables('test')");
Based purely on my knowledge of regex (and not how they are done in C#), the pattern you want is probably:
"[^"]*"
ie - match a " then match everything that's not a " then match another "
You may need to escape the double-quotes to make your regex-parser actually match on them... that's what I don't know about C#
Try to avoid where you can the '.*' in regex, you can usually find what you want to get by avoiding other characters, for example [^"]+ not quoted, or ([^)]+) not in parenthesis. So you may just want "([^"]+)" which should give you the whole thing in [0], then in [1] you'll find 'test'.
You could also just replace '"' with '' I think.
Taryn Easts regex includes the *. You should remove it, if it is just a placeholder for any value:
"[^"]"
BTW: You can test this regex with this cool editor: http://rubular.com/r/1MMtJNF3kM
hi , I have 2 related questions.
1)suppose we have:
string strMessage="\nHellow\n\nWorld";
console.writeln(strMessage);
Result is:
Hellow
World
Now if we want to show the string in the original format in One Line
we must redefine the first variable from scratch.
string strOrignelMessage=#"\nHellow\n\nWorld" ;
console.writln(strOrignelMessage);
Result is:
\nHellow\n\nWorld --------------------->and everything is ok.
i am wondering is there a way to avoid definning
the new variable(strOrignelMessage) in code for this purpose and just using only
the first string variable(strMessage) and apply some tricks and print it in one line.
at first i tried the following workaround but it makes some bugs.suppose we have:
string strMessage="a\aa\nbb\nc\rccc";
string strOrigenalMessage=strMessage.replace("\n","\\n").replace("\r","\\r");
Console.writeln(strOrigenalMessage)
result is :aa\nbb\nc\rccc
notice that befor the first "\" not printed.and now my second question is:
2)How we can fix the new problem with single "\"in the string
i hope to entitle this issue correctly and my explanations would be enough,thanks
No, because the compiler has already converted all of your escaped characters in the original string to the characters they represent. After the fact, it is too late to convert them to non-special characters. You can do a search and replace, converting '\n' to literally #"\n", but that is whacky and you're better off defining the string correctly in the first place. If you wanted to escape the backslashes in the first place, why not put an extra backslash character in front of each of them:
Instead of "\n" use "\\n".
Updated in response to your comment:
If the string is coming from user input, you don't need to escape the backslash, because it will be stored as a backslash in the input string. The escape character only works as an escape character in string literals in code (and not preceded by #, which makes them verbatim string literals).
if you want "\n\n\a\a\r\blah" to print as \n\n\a\a\r\blah without # just replace all \ with \\
\ is the escaper in a non-verbatim string. So you simply need to escape the escaper, as it were.
If you want to use both strings, but want to have only one in the code then write the string with #, and construct the other one with Replace(#"\n","\n").
explanations for Anthony Pegram (if i understand u right) and anyone that found it usefull
i think i find my way in question2.
at first ,unfortunately,i thought that the
escape characters limts to \n,\t,\r,\v and
this made me confuesed becouse in my sample string i used \a and \b
and the compiler behaviuor was not understandable for me.
but finally i found that \a and \b is in
escape-characters set too.and if u use "\" without escap characters
a compile time error would be raised (its so funny when i think to My mistake again)
pls refers to this usefull msdn article for more info.
2.4.4.5 String literals
and you couldnt replace \ (single\) with \\
becouse fundamentally you couldnt have a (single \) without using
escape-characters after it in a string .so we coudnt write such a string in the code:
string strTest="abc\pwww"; ------> compile time error
and for retriving an inactived escape characters version of a string
we can use simply string.replace method as i used befor.
excuse me for long strory ,thank u all for cooperation.