${str?replace("\d+", "", "r")};
I wanted to use \d to remove numbers, but it didn't work!!!
But ${str?replace("[0-9]", "", "r")}; works!!!
So, I wanna know how to use regex like \d, \b, \w, etc?
You need to double the backslashes:
${str?replace("\\d+", "", "r")};
This is because string escaping rules are applied before regex escaping rules. So the string "\\d" is translated to the regex \d which then matches a digit.
If your string is "\d", the string processor translates it to a literal d (because \d is not a recognized string escape sequence, so it's ignored).
Related
As you probably all know, regular expressions have some metacharacters, such as \, |, ., ?, +, *,…. If you want to search for a substring including one of these characters without actually using the regex behaviour, you can escape it with a backslash.
So if you want to search for "Is it true?" in a string you would use the pattern
"Is it true\?".
I am using Kotlin and its built-in regular expressions. Is there a way in Kotlin (a function or something) to get a string from another string in which all of the special characters in the input string are escaped?
So if the input to such a function were "This is good." the output would be "This is good\.", and for "? a [+" it would be "\? a \[\+". → every regex special character in the output is escaped with a backslash.
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 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.
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.
Within an ERE, a backslash character (\, \a, \b, \f, \n,
\r, \t, \v) is considered to begin an escape sequence.
Then I see \\n and [\\\n], I can guess though both \\n and [\\\n] here means \ followed by new line, but I'm confused by the exact process to interpret such sequence as how many \s are required at all?
UPDATE
I don't have problem understanding regex in programing languages so please make the context within the lexer.
[root# ]# echo "test\
> hi"
This is dependent on the programming language and on its string handling options.
For example, in Java strings, if you need a literal backslash in a string, you need to double it. So the regex \n must be written as "\\n". If you plan to match a backslash using a regex, then you need to escape it twice - once for Java's string handler, and once for the regex engine. So, to match \, the regex is \\, and the corresponding Java string is "\\\\".
Many programming languages have special "verbatim" or "raw" strings where you don't need to escape backslashes. So the regex \n can be written as a normal Python string as "\\n" or as a Python raw string as r"\n". The Python string "\n" is the actual newline character.
This can becoming confusing, because sometimes not escaping the backslash happens to work. For example the Python string "\d\n" happens to work as a regex that's intended to match a digit, followed by a newline. This is because \d isn't a recognized character escape sequence in Python strings, so it's kept as a literal \d and fed that way to the regex engine. The \n is translated to an actual newline, but that happens to match the newline in the string that the regex is tested against.
However, if you forget to escape a backslash where the resulting sequence is a valid character escape sequence, bad things happen. For example, the regex \bfoo\b matches an entire word foo (but it doesn't match the foo in foobar). If you write the regex string as "\bfoo\b", the \bs are translated into backspace characters by the string processor, so the regex engine is told to match <backspace>foo<backspace> which obviously will fail.
Solution: Always use verbatim strings where you have them (e. g. Python's r"...", .NET's #"...") or use regex literals where you have them (e. g. JavaScript's and Ruby's /.../). Or use RegexBuddy to automatically translate the regex for you into your language's special format.
To get back to your examples:
\\n as a regex means "Match a backslash, followed by n"
[\\\n] as a regex means "Match either a backslash or a newline character".
Actually regex string specified by string literal is processed by two compilers: programming language compiler and regexp compiler:
Original Compiled Regex compiled
"\n" NL NL
"\\n" '\'+'n' NL
"\\\n" '\'+NL NL
"\\\\n" '\'+'\'+'n' '\'+'n'
So you must use the shortest format "\n".
Code examples:
JavaScript:
'a\nb'.replace(RegExp("\n"),'<br>')
'a\nb'.replace(RegExp("\\n"),'<br>')
'a\nb'.replace(RegExp("\\\n"),'<br>')
but not:
'a\nb'.replace(/\\\n/,'<br>')
Java:
System.out.println("a\nb".replaceAll("\n","<br>"));
System.out.println("a\nb".replaceAll("\\n","<br>"));
System.out.println("a\nb".replaceAll("\\\n","<br>"));
Python:
str.join('<br>',regex.split('\n','a\nb'))
str.join('<br>',regex.split('\\n','a\nb'))
str.join('<br>',regex.split('\\\n','a\nb'))