Ignoring bracketed sections in a regex match [closed] - regex

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Right now I have this regex:
/regular(\[[a-z]*\])* expression/i
which correctly matches this string:
Regular[AECC][XVK] Expression
but fails to match these:
R[AECC]egular [XVK]Expression
R[ABC]egular Expression
Regular[xx] Expressio[]n
How can I match all of the above with a single regex?


/
r (?:\[[a-z]*\])?
e (?:\[[a-z]*\])?
g (?:\[[a-z]*\])?
...
i (?:\[[a-z]*\])?
o (?:\[[a-z]*\])?
n
/ix
or
my $pat = join '(?:\\[[a-z]*\\])?', map quotemeta, split //, 'regular expression';
my $re = qr/$pat/i;
/$re/
or
s/\[[a-z]*\]//ig;
/regular expression/i

Related

Split string by word contain space in Java [closed]

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I have a String pattern: "hh:mm:ss dd:MM:yyyy to hh:mm:ss dd:MM:yyyy" and I want to extract date String from it.
Example:
S = "00:00:00 19/08/2022 to 23:59:59 19/08/2022"
Split into S1 = "00:00:00 19/08/2022" and S2 = "23:59:59 19/08/2022".
I'm trying to use String.split function but can't figure out the regex yet. Can somebody help?
I'm using Java 8.

Just split on \s+to\s+:
String pattern = "00:00:00 19/08/2022 to 23:59:59 19/08/2022";
String[] parts = pattern.split("\\s+to\\s+");
System.out.println(Arrays.toString(parts));
This prints:
[00:00:00 19/08/2022, 23:59:59 19/08/2022]

Find and Replace , between 2 characters [closed]

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So I have a CSV file with some anomaly for eg
2019-07-25 00:00:00,1014488,2019-07-25 12:24:12,112629,Amy,Flutmus,84004,GM,0001,2.99,312,FFO & CS PLATE ||22,10999,90027,90062||Sand w/ Options,1,0,0.2,18.85,0,1
i want to replace , between these characters || ||.
So I'm expecting
2019-07-25 00:00:00,1014488,2019-07-25 12:24:12,112629,Amy,Flutmus,84004,GM,0001,2.99,312,FFO & CS PLATE ,22,*10999*90027*90062,Sand w/ Options,1,0,0.2,18.85,0,1

You could use re.sub to capture all your strings between || and then replacing the ,s with *s:
import re
value = "2019-07-25 00:00:00,1014488,2019-07-25 12:24:12,112629,Amy,Flutmus,84004,GM,0001,2.99,312,FFO & CS PLATE ||22,10999,90027,90062||Sand w/ Options,1,0,0.2,18.85,0,1"
pattern = re.compile(r'\|\|(.+)\|\|')
cleaned_value = pattern.sub(lambda match: match.group().replace(",", "*"), value)
print(cleaned_value.replace(r'||', ','))

Extract range using regular expression [closed]

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For example
1cn1
1cn2
1cn3
1cn4
1cn5
1cn6
1cn7
1cn8
1cn9
1cn10
1cn11
1cn12
extract lines between 1cn8 to 1cn12
like this i have hundreds of line, want to extract any range by giving the input.
$ cat test | grep '[$0-9]'
1cn1
1cn2
1cn3
1cn4
1cn5
1cn6
1cn7
1cn8
1cn9
1cn10
1cn11
1cn12
these are node names, want to extract node names within the node range. lines starting from1cn8 till 1cn12

You can try something like that:
[[:alnum:]]{3}([8-9]|1[0-2])
Short explanation:
[[:alnum:]]{3}: matchs with any letter (upper or lower case) and digits, over 3 times;
([8-9]|1[0-2]): matchs with 8 and 9 [8-9] OR |, the number one with 0, 1 or 2, 1[0-2]
$ cat test
1cn1
1cn2
1cn3
1cn4
1cn5
1cn6
1cn7
1cn8
1cn9
1cn10
1cn11
1cn12
$ egrep '[[:alnum:]]{3}([8-9]|1[0-2])' test
1cn8
1cn9
1cn10
1cn11
1cn12

I have a string in perl, how do I replace it with the same string but with [ ] around only the numbers? [closed]

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Example: DATA500 replaced with DATA[500]

This should work. It replaces groups of one or more digits, the "\d+" part, with the captured string surrounded by [], the "[$1]" part.
$a = "bob123bob123";
$a =~ s/(\d+)/[$1]/g;
print "$a";
# bob[123]bob[123]

Regular expression which except a-z , A-Z ,0-9 and -(dash) [closed]

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I need write a regular expression which accept a-z , A-Z,0-9 and -(dash) as a special character .
valid inputs : 1 : mohit-kumar-gulati
2 : mohit1-kumar1-gulati
3 : mohit1-kumar1-gulati
4 : 234-545-345
Invalid inputs : 1 : mohit#-kumar-gulati
2 : mohit1-kumar1$-gulati
3 : ###%^-kumar1-gulati
4 : %^-::-''::
5 : *(*mohit

Use this one:
[A-Za-z]*-?[A-Za-z]*

If it is C# you should use a regex like this:
Regex regex = new Regex(#"^[-\w]+$");
Good luck with your quest.