I want to replace dots with commas for some but not all matches:
hostname_metric (Index: 1) to hostname;metric (avg);22.04.2015 13:40:00;3.0000;22.04.2015 02:05:00;2.0000;22.04.2015 02:00:00;650.7000;2.2594;
The outcome should look like this:
hostname_metric (Index: 1) to hostname;metric (avg);22.04.2015 13:40:00;3,0000;22.04.2015 02:05:00;2,0000;22.04.2015 02:00:00;650,7000;2,2594;
I was able to identify the RegEx which should work to find the correct dots.
;[0-9]{1,}\.[0-9]{4}
But how can I replace them with a comma with awk or sed?
Thanks in advance!
Adding some capture groups to the regex in your question, you can use this sed one-liner:
sed -r 's/(;[0-9]{1,})\.([0-9]{4})/\1,\2/g' file
This matches and captures the part before and after the . and uses them in the replacement string.
On some versions of sed, you may need to use -E instead of -r to enable Extended Regular Expressions. If your version of sed doesn't understand either switch, you can use basic regular expressions and add a few escape characters:
sed 's/\(;[0-9]\{1,\}\)\.\([0-9]\{4\}\)/\1,\2/g' file
sed 's/\(;[0-9]\+\)\.\([0-9]\{4\}\)/\1,\2/g' should do the trick.
Related
I want to search recursiv in files for a given pattern and replace them. The search is for a string like "['DB']['1']['HOST'] = 'localhost'". If testing the regex the following doesn't print anything. Can't see an error in this regex? Could anyone help?
sed -n '/\[\'HOST\'\]\s?=\s?(?:\'|")(.+)(?:\'|")/p' /path/to/file
POSIX regex does not support non-capturing groups. Besides, you have not specified the -E option and the pattern is parsed as a BRE POSIX pattern where the capturing parentheses should be escaped. Also, the single quotes cannot be escaped to be used in a sed regex pattern, use \x27 instead.
Use
sed -En '/\[\x27HOST\x27\]\s?=\s?[\x27"][^\x27"]+[\x27"]/p'
See an online demo:
s="a string like ['DB']['1']['HOST'] = 'localhost'."
sed -En '/\[\x27HOST\x27\]\s?=\s?[\x27"][^\x27"]+[\x27"]/p' <<< "$s"
Besides, instead of \s, it might be a good idea to use [[:space:]].
So I'm trying to match a regexp with any string in the middle of it and then print out just that string. The syntax is sort of like this...
sed -n 's/<title>.*</title>/"what do I put here"/p' input.file
and I just want to print out whatever .* is where I typed "what do I put here". I'm not very comfortable with sed at this point so this is likely a very simple answer and I'm having trouble finding one in any of the other questions. Thanks in advance!
Capture the pattern you want to extract within \(...\), and then you can refer to it as \1 in the replacement string:
sed -n 's/<title>\(.*\)</title>/\1/p' input.file
You can have multiple \(...\) expressions, and refer to them with \1, \2, \3, and so on.
If you have the GNU version of sed, or gsed, then you could simplify a bit:
sed -rn 's/<title>(.*)</title>/\1/p' input.file
With the -r flag, sed can use "extended regular expressions", which practically let's you write (...) instead of \(...\), + instead of \+, and other goodies.
I need to use sed to look for all lines in a file with pattern "[whatever]|[whatever]" so I'm using the following regex:
sed '/\"[a-zA-Z0-9]+\|[a-zA-Z0-9]+\"/p' test2.txt
But it's not working because in this file is returning something when it shouldn't
RTV0031605951US|3160595|20/03/2013|0|"Laurie Graham"|"401"
Does anybody know with regex should I use? Thanks in advance
I see three problems with your regular expression:
+ is not a metacharacter, so you need to escape it to get its special meaning.
Similar issue happens with the pipe. Neither it is a metacharacter, so don't escape it to match it literally.
Sed by default prints each line that matches, so add -n that avoids that, if you already use /p that prints it. Otherwise you will have those lines twice in the output.
sed will output anything that is a partial match.
To match only whole lines that match your regex, add ^ and $ to the start/end:
sed '/^\"[a-zA-Z0-9]+\|[a-zA-Z0-9]+\"$/p' test2.txt
sed '/\B\"[ [:alnum:]]\+\"|\"[ [:alnum:]]\+\"\B/!d' file
If you use this in a sed script, do not escape double quotes.
I have the following strings in a text file (big one, more like these and different):
79A18D7F-1517-5981-8446-3A0452727B06
7842A72D-1517-5281-84E4-EAEF09B743F7
6040BEE7-1517-5982-84C1-419B224E647E
615F2747-1517-5981-84AF-787C34967FB2
7468A3E3-1517-5931-84B3-3FC3F701C269
I can find them using grep and regex:
'[0-9A-F]{8}-[0-9]{4}-[0-9]{4}-[0-9A-F]{4}-[0-9A-F]{12}'
what's the sed regex syntax to delete them because:
sed "s/[0-9A-F]{8}-[0-9]{4}-[0-9]{4}-[0-9A-F]{4}-[0-9A-F]{12}//g"
doesn't seem to work.
Thanks!
Use sed -r. You are relying on extended regular expression syntax features without escaping them, but with sed -r you don't have to. If you want to actually delete the lines instead of just clearing them, you can use:
sed -r "/regex/d"
In addition, for regular sed (BRE) you would need to escape the curly braces:
sed 's/[0-9A-F]\{8\}-[0-9]\{4\}-[0-9]\{4\}-[0-9A-F]\{4\}-[0-9A-F]\{12\}//g' file
I need to find out how to delete up to 10 digits that are at the end of the line in my text file using sed.
For example if I have this:
ajsdlfkjasldf1234567890
asdlkjfalskdjf123456
adsf;lkjasldfkjas123
it should become:
ajsdlfkjasldf
asdlkjfalskdjf
adsf;lkjasldfkjas
can anyone help?
I have this, but its not working:
sed 's/[0-9]{10}$//g'
Have you tried this:
sed 's/[0-9]+$//'
Your command would only match and delete exactly 10 digits at the end of line and only, if you enabled extended regular expressions (-E or -r, depending on your version of sed).
You should try
sed -r 's/[0-9]{1,10}$//'
The following should work:
sed 's/[0-9]\{1,10\}$//' file
Regex syntax in sed requires backslashes before the brackets to use them for repetition, unless you use an extended regex option.
A quick look here suggests you should try this:
$ sed 's/[0-9]\{0,10\}$//g'
{ } should be escaped, unless you switch to extended regex syntax:
$ sed -r 's/[0-9]{0,10}$//g'