Regex match backslash star - regex

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 "\\\\\\*".

Related

How to exclude part of string using regex and change add this part and the and of string?

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.)

How to prepend a backslash using clojure.string/replace

I am trying to use clojure.string/replace to escape certain characters like asterisks and backticks with backslashes (like ex*mple -> ex\*mple), but I cannot make sense of the function's own escaping rules:
If I try (cs/replace "ex*mple" #"[\*`]" "\\$0"), it treats the $0 literally and returns ex$0mple.
If I try (cs/replace "ex*mple" #"[\*`]" "\\\\$0") it adds two slashes: ex\\*mple.
What is the right way to do it?
Your second approach, (cs/replace "ex*mple" #"[\*`]" "\\\\$0"), is correct. The reason you see two backslashes in the result is because that's how Clojure shows single backslashes in strings. If you print "ex\\*mple", you'll see ex\*mple.
Clojure uses backslash as an escape character in strings, so backslashes themselves have to be escaped. ex\*mple is not a valid string in Clojure because \* is an unsupported escape character.

Regex replace escaped characters

I would like to define a regex pattern which replaces escaped characters with the corresponding value.
For example the string
xy\tz\\x
Should be converted to
xy{tab}z\x
The problem is how to handle things like
xy\\\\\t
this string should become
xy\\{tab}
I don't know how to create a pattern which matches only odd backslashes.
This isn't something that can be accomplished using a single pattern. To start, strip out collections of backslashes:
s/\\\\/\\/g
This replaces two backslashes with a single one.
Then you can just apply one pattern per escaped character:
s/\\t/\t/g
The trick here is to escape the backslash you want to replace. What this'll do is replace the literal string "\t" with a tab character.

Regex Everything except !"#$'*+,/:;\`|

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).

Visual Studio Find and Replace with regex and single/double quotes

How to use VS Find/Replace to replace:
this: $('a[name="lnkFind"]').on('click', function
with this: $(document).on("click", "a[name='lnkFind']", function
I'm not sure which characters need to be escaped - single or double quotes or both? None of the patters I've tried seem to find a match.
You'll need to escape many of these characters.
Find/Replace will complain about the un-escaped ( and ), even the bare ( at the end because it's missing a matching ). Also the square brackets, which are used for character sets, and finally the $.
So this should work as the pattern:
\$\('a\[name="lnkFind"\]'\).on\('click', function
You should look at a list of special characters in Regular Expressions.
$, ., [, ] should all be escaped.
http://www.fon.hum.uva.nl/praat/manual/Regular_expressions_1__Special_characters.html
Except in special cases (such as vim regex), in general you can escape any and all special characters in regex to get their literal form, i.e. escaping a special character that doesn't need to be escaped, won't do any harm.
That said, here's the minimum that needs to be escaped:
\$\('a\[name="lnkFind"]')\.on\('click', function
I don't think you'll need to escape anything in the replacement, because only a $ or \ followed by a number will be interpreted.