How to match (without capturing): 5678
In this string: performer=5678,25678,56342,56782
This is what I have so far: 5678(?!\d)
http://rubular.com/r/WqdK6sQjOK
\b5678\b
... which works also if you delimit your numbers with other separators, such as ;, /, etc
Related
How to get a regular expression to replace all characters in a string in perl with *? The string has some utf-8 or iso-8859-1 characters also. I tried with "s/\w/*/g". But it did not replace utf-8 or iso-8859-1 characters.
my $value="hellö";
print "$value\n";
$value =~ s/\w/*/g;
print "after replacing $value\n"; //It prints ****ö.
I expect all characters should be replaced with * i.e hellö should be replaced with *****.
Please note, few special characters like -,_,\,/ etc should be skipped.
If you want to skip just a few characters, you can always do something along the lines of
s/[^, \/\\\-]/*/g;
To replace all the characters in a string? The \w is for matching word characters, but using just a dot should match all characters: s/./*/g
So i have an alphanumeric string 10006cc2190ab011 i am trying to add a colon after every two letters in this alphanumeric string.
this is the string : 10006cc2190ab011
i want it be - 10:00:6c:c2:19:0a:b0:11
Thanks in advance.
A sed solution:
$ echo 10006cc2190ab011 | sed 's/../&:/g; s/:$//'
10:00:6c:c2:19:0a:b0:11
Replaces each non-overlapping pair of characters with the same pair plus :. In the end removes the trailing : (if input text had even length).
str=10006cc2190ab011; str="${str//??/${.sh.match}:}"; echo ${str%:}
is doing the same replacement without the use of an external command, just using ksh-internals.
Doing the same as in sed (the other answer). Replace in $str every // two charactes ?? with / the matched string and a : (every match is kept in the ksh-variable ${.sh.match}). Then print $str without the last % ':'.
My string:
a = "Please match spaces here <but not here>. Again match here <while ignoring these>"
Using Ruby's regex flavor, I would like to do something like:
a.gsub /regex_pattern/, '_'
And obtain:
"Please_match_spaces_here_<but not here>._Again_match_here_<while ignoring these>"
This should do it:
result = subject.gsub(/\s+(?![^<>]*>)/, '_')
This regex assumes there's nothing tricky like escaped angle brackets. Also be aware that \s matches newlines, TABs and other whitespace characters as well as spaces. That's probably what you want, but you have the option of matching only spaces:
/ +(?![^<>]*>)/
I think, it works:
a = "Please match spaces here <but not here>. Again match here <while ignoring these>"
pattern = /<(?:(?!<).)*>/
a.gsub(pattern, '')
# => "Please match spaces here . Again match here "
Given the following inputs:
line1 = "Hey | Hello | Good | Morning"
line2 = "Hey , Hello , Good , Morning"
file1=length1=name1=title1=nil
Using ',' to split the string as follows:
file1, length1, name1, title1 = line2.split(/,\s*/)
I get the following output:
puts file1,length1,name1,title1
>Hey
>Hello
>Good
>Morning
However, using '|' to split the string I receive a different output:
file1, length1, name1, title1 = line2.split(/|\s*/)
puts file1,length1,name1,title1
>H
>e
>y
Both the strings are same except the separating symbol (a comma in first case and a pipe in second case). The format of the split function I am using is also the same except, of course, for the delimiting character. What causes this variation?
The problem is because | has the meaning of OR in regex. If you want literal character, then you need to escape it \|. So the correct regex should be /\|\s*/
Currently, the regex /|\s*/ means empty string or series of whitespace character. Since the empty string specified first in the OR, the regex engine will break the string up at every character (you can imagine that there is an empty string between characters). If you swap it to /\s*|/, then the whitespaces will be preferred over empty string where possible and there will be no white spaces in the list of tokens after splitting.
What is the regex to match a string with at least one period and no spaces?
You can use this :
/^\S*\.\S*$/
It works like this :
^ <-- Starts with
\S <-- Any character but white spaces (notice the upper case) (same as [^ \t\r\n])
* <-- Repeated but not mandatory
\. <-- A period
\S <-- Any character but white spaces
* <-- Repeated but not mandatory
$ <-- Ends here
You can replace \S by [^ ] to work strictly with spaces (not with tabs etc.)
Something like
^[^ ]*\.[^ ]*$
(match any non-spaces, then a period, then some more non-spaces)
no need regular expression. Keep it simple
>> s="test.txt"
=> "test.txt"
>> s["."] and s.count(" ")<1
=> true
>> s="test with spaces.txt"
=> "test with spaces.txt"
>> s["."] and s.count(" ")<1
=> false
Try this:
/^\S*\.\S*$/