I want every character possible (letters,numbers,everyting) except: !"#$'*+,/:;\|`
Is this correct? [^!"#$'*+,/:;\`|]
I made it work using http://www.regextester.com/ and your suggestions
*WORKS--> [^!"#$'*+,/:;\\`|]
A regex pattern to match a single \:
\\
It means you need to escape a backslash by adding another backslash to it. So, you need to use
[^!"#$'*+,/:;\\`|]
^^
Note that if you use it in a regular string literal in some programming language, you will have to double each of the backslashes (so that there are 4 backslashes).
Related
I've got a little problem with regex.
I got few strings in one file looking like this:
TEST.SYSCOP01.D%%ODATE
TEST.SYSCOP02.D%%ODATE
TEST.SYSCOP03.D%%ODATE
...
What I need is to define correct regex and change those string name for:
TEST.D%%ODATE.SYSCOP.#01
TEST.D%%ODATE.SYSCOP.#02
TEST.D%%ODATE.SYSCOP.#03
Actually, I got my regex:
r".SYSCOP[0-9]{2}.D%%ODATE" - for finding this in file
But how should look like the changing regex? I need to have the numbers from a string at the and of new string name.
.D%%ODATE.SYSCOP.# - this is just string, no regex and It didn't work
Any idea?
Find: (SYSCOP)(\d+)\.(D%%ODATE)
Replace: $3.$1.#$2 or \3.\1.#\2 for Python
Demo
You may use capturing groups with backreferences in the replacement part:
s = re.sub(r'(\.SYSCOP)([0-9]{2})(\.D%%ODATE)', r'\3\1.#\2', s)
See the regex demo
Each \X in the replacement pattern refers to the Nth parentheses in the pattern, thus, you may rearrange the match value as per your needs.
Note that . must be escaped to match a literal dot.
Please mind the raw string literal, the r prefix before the string literals helps you avoid excessive backslashes. '\3\1.#\2' is not the same as r'\3\1.#\2', you may print the string literals and see for yourself. In short, inside raw string literals, string escape sequences like \a, \f, \n or \r are not recognized, and the backslash is treated as a literal backslash, just the one that is used to build regex escape sequences (note that r'\n' and '\n' both match a newline since the first one is a regex escape sequence matching a newline and the second is a literal LF symbol.)
I want to detect a double quote enclosed string with a regular expression.
I did this with a simple: /"[^"]*"/
But this doesn't allow for double quotes inside the string, so I would ideally allow double quote by escaping it like \", and of course that would need also allowing \\ for a single backslash or else a string could never end in a backslash.
How can I accomplish this? ideally with a Perl-compatible regular expression.
PS: No need to escape single quotes and not possible to use them to enclose the string.
Here you go:
"(?:\\.|[^"])*"
Demo
For each character in the string, match either a backslash followed by anything, or a character that is not a quote.
And if you need something optimized, here's an alternative:
"(?>[^\\"]++|\\.)*+"
Demo
It basically uses possessive quantifiers to avoid backtracking.
I'm trying to match a variable length string followed by the filetype suffix in an XML filename using a regex:
varrrrrriableLengthString.xml
Currently I'm using this regex with a greedy match, the second backslash is to escape the first, which is to escape the dot.
[A-Za-z0-9]+\\.[xX][mM][lL]
I've tested this on RegExr, and it matches with only one backslash. However my CPP parser requires the double backslash.
How can I properly escape the filename suffix?
You can also escape chars using the [] notation, in your case [.]. The main advantage is that there is no "one or two backslashes?" question anymore, and I find it more readable IMHO.
It just does not work with brackets, i.e. to escape a [ (or ]), you still have to use \[ (or \\[ for a string literal) and not [[].
Backslashes still have to be escaped using another backslash too.
Can't work this one out, this matches a single star:
// Escaped multiply
Text = Text.replace(new RegExp("\\*", "g"), '[MULTIPLY]');
But I need it to match \*, I've tried:
\\*
\\\\*
\\\\\*
Can't work it out, thanks for any help!
You were close, \\\\\\* would have done it.
Better use verbatim strings, that makes it easier:
RegExp(#"\\\*", "g")
\\ matches a literal backslash (\\\\ in a normal string), \* matches an asterisk (\\* in a normal string).
Remember that there are two 'levels' of escaping.
First, you are escaping your strings for the C# compiler, and you are also escaping your strings for the Regex engine.
If you want to match "\*" literally, then you need to escape both of these characters for the regex engine, since otherwise they mean something different. We escape these with backslashes, so you will have "\\\*".
Then, we have to escape the backslashes in order to write them as a literal string. This means replacing each backslash with two backslashes: "\\\\\\*".
Instead of this last part, we could use a "verbatim string", where no escapes are applied. In this case, you only need the result from the first escaping: #"\\\*".
Your syntax is completely wrong. It looks more like Javascript than C#.
This works fine:
string Text = "asdf*sadf";
Text = Regex.Replace(Text, "\\*", "[MULTIPLY]");
Console.WriteLine(Text);
Output:
asdf[MULTIPLY]sadf
To match \* you would use the pattern "\\\\\\*".
I need a Perl regular expression to match a string. I'm assuming only double-quoted strings, that a \" is a literal quote character and NOT the end of the string, and that a \ is a literal backslash character and should not escape a quote character. If it's not clear, some examples:
"\"" # string is 1 character long, contains dobule quote
"\\" # string is 1 character long, contains backslash
"\\\"" # string is 2 characters long, contains backslash and double quote
"\\\\" # string is 2 characters long, contains two backslashes
I need a regular expression that can recognize all 4 of these possibilities, and all other simple variations on those possibilities, as valid strings. What I have now is:
/".*[^\\]"/
But that's not right - it won't match any of those except the first one. Can anyone give me a push in the right direction on how to handle this?
/"(?:[^\\"]|\\.)*"/
This is almost the same as Cal's answer, but has the advantage of matching strings containing escape codes such as \n.
The ?: characters are there to prevent the contained expression being saved as a backreference, but they can be removed.
NOTE: as pointed out by Louis Semprini, this is limited to 32kb texts due a recursion limit built into Perl's regex engine (that unfortunately silently returns a failure when hit, instead of crashing loudly).
How about this?
/"([^\\"]|\\\\|\\")*"/
matches zero or more characters that aren't slashes or quotes OR two slashes OR a slash then a quote
A generic solution(matching all backslashed characters):
/ \A " # Start of string and opening quote
(?: # Start group
[^\\"] # Anything but a backslash or a quote
| # or
\\. # Backslash and anything
)* # End of group
" \z # Closing quote and end of string
/xms
See Text::Balanced. It's better than reinvent wheel. Use gen_delimited_pat to see result pattern and learn form it.
RegExp::Common is another useful tool to be aware of. It contains regexps for many common cases, included quoted strings:
use Regexp::Common;
my $str = '" this is a \" quoted string"';
if ($str =~ $RE{quoted}) {
# do something
}
Here's a very simple way:
/"(?:\\?.)*?"/
Just remember if you're embedding such a regex in a string to double the backslashes.
Try this piece of code : (\".+")