I have function that prints out the longest path in directory tree. Let's say the function prints this: ./.mozilla/firefox/z6upkljn.default/storage/permanent/chrome/idb/2918063365piupsah.files
What I want to do is to cut this string after match with user defined regex.
For example if user puts in regex like: *de?a*, the only match is z6upkljn.default. So at the end, the output will be ./.mozilla/firefox
Here is a code sample I found sed 's/My_expression.*//'
Where the My_expression is regular expression and delimiter for cutting defined by user.
It works for this input $echo /homes/eva/xm/xmikfi00 | sed 's/mik.*//', where for output I get /homes/eva/xm/x. As expected.
But if I enter simple regex $echo /homes/eva/xm/xmifki00 | sed 's/mi?.*//', the output is /homes/eva/xm/xmikfi00. Anyone who can help me how to get the same output as in the previous example?
I'll be glad for any help or suggestions, thanks.
Sed uses (by default) POSIX BREs, not EREs. If what you're trying to match with your ? is "any character", use a .: echo /homes/eva/xm/xmifki00 | sed 's/mi..*//'.
See man 7 re_format for more details.
Related
I am trying to write a script which can extract the words from a string untill the first number appears.
ex :- I have a file named as typed-list-4.1.3.Final.jar and I want the output as:- typed-list.jar
Since all the files have different names, but, they end with a version number and .jar extension so I was trying to sed the part from where the first number appears and then append .jar.
My files look like :-
log4j-slf4j-impl-2.8.2.jar, hibernate-core-5.0.12.Final.jar etc
I tried to use sed command like this but it's not working :-
sed -i 's/-[0-9]*$//g' test1.sh --- where test1.sh contains this string "typed-list-4.1.3.Final.jar"
How about:
sed 's/-\([0-9]\+\.\)\+[0-9]\+.*\.jar/.jar/' Input_file
Results for the provided inputs:
typed-list.jar
log4j-slf4j-impl.jar
hibernate-core.jar
The regex matches with a substring such as:
starting with a dash -
pattern repetition of digit(s) dot digit(s) ...
some other substring in between (such as Final)
ends with the extension .jar
Then the sed command replaces the matched substring with just the extension.
Hope this helps.
Sed:
sed -E 's/(.*)-([[:digit:]]+\.){2}[[:digit:]]+.*(\.[^.]+)$/\1\3/' dat
log4j-slf4j-impl.jar
hibernate-core.jar
typed-list.jar
echo typed-list-4.1.3.Final.jar | awk 'sub(/-4.{10}/,"",$0)'
typed-list.jar
I have an issue with string manipulation in bash. I have a list of names, each name being composed of two parts, chars and numbers: for example
abcdef01234
I want to cut the last character before the numeric part starts, in this case
f
I think there is a regular expression to help me with this but just can't figure it out. AWK/sed solutions are accepted too. Hope someone can help.
Thank you.
In bash it can be done with parameter expansion with substring removal and string indexes, e.g.,
a=abcdef01234 # your string
tmp=${a%%[0-9]*} # remove all numbers from right
echo ${tmp:(-1)} # output last of remaining chars
Output: f
You can use a regexp like [a-zA-Z]+([a-zA-Z])[0-9]+. If you know how to use sed is pretty easy.
Check https://regex101.com/r/XCkKM5/1
The match will be the letter you want.
^\w+([a-zA-Z])\d+$
As a sed command (on OSX) this will be :
echo "abcdef12345" | sed -E "s#^[a-zA-Z]+([a-zA-Z])[0-9]+\$#\1#"
try following too once.
echo "abcdef01234" | awk '{match($0,/[a-zA-Z]+/);print substr($0,RLENGTH,1)}'
I have a list of names I assume is a file, file. Using grep's PCRE and (positive) lookahead:
$ grep -oP "[a-z](?=[^a-z])" file
f
It prints out the first (lowercase) letter followed by a non-(lowercase)-letter.
I want to separate string and number in a file to get a specific number in bash script, such as:
Branches executed:75.38% of 1190
I want to only get number
75.38
. I have try like the code below
$new_value=value | sed -r 's/.*_([0-9]*)\..*/\1/g'
but it was incorrect and it was failed.
How should it works? Thank you before for your help.
You can use the following regex to extract the first number in a line:
^[^0-9]*\([0-9.]*\).*$
Usage:
% echo 'Branches executed:75.38% of 1190' | sed 's/^[^0-9]*\([0-9.]*\).*$/\1/'
75.38
Give this a try:
value=$(sed "s/^Branches executed:\([0-9][.0-9]*[0-9]*\)%.*$/\1/" afile)
It is assumed that the line appears only once in afile.
The value is stored in the value variable.
There are several things here that we could improve. One is that you need to escape the parentheses in sed: \(...\)
Another one is that it would be good to have a full specification of the input strings as well as a good script that can help us to play with this.
Anyway, this is my first attempt:
Update: I added a little more bash around this regex so it'll be more easy to play with it:
value='Branches executed:75.38% of 1190'
new_value=`echo $value | sed -e 's/[^0-9]*\([0-9]*\.[0-9]*\).*/\1/g'`
echo $new_value
Update 2: as john pointed out, it will match only numbers that contain a decimal dot. We can fix it with an optional group: \(\.[0-9]\+\)?.
An explanation for the optional group:
\(...\) is a group.
\(...\)? Is a group that appears zero or one times (mind the question mark).
\.[0-9]\+ is the pattern for a dot and one or more digits.
Putting all together:
value='Branches executed:75.38% of 1190'
new_value=`echo $value | sed -e 's/[^0-9]*\([0-9]\+\(\.[0-9]\+\)\?\).*/\1/g'`
echo $new_value
Given a string "pos:665181533 pts:11360 t:11.360000 crop=720:568:0:4 some more words"
Is it possible to extract string between "crop=" and the following space using bash and grep?
So if I match "crop=" how can I extract anything after it and before the following white space?
Basically, I need "720:568:0:4" to be printed.
I'd do it this way:
grep -o -E 'crop=[^ ]+' | sed 's/crop=//'
It uses sed which is also a standard command. You can, of course, replace it with another sequence of greps, but only if it's really needed.
I would use sed as follows:
echo "pos:665181533 pts:11360 t:11.360000 crop=720:568:0:4 some more words" | sed 's/.*crop=\([0-9.:]*\)\(.*\)/\1/'
Explanation:
s/ : substitute
.*crop= : everything up to and including "crop="
\([0-9.:]\) : match only numbers and '.' and ':' - I call this the backslash-bracketed expression
\(.*\) : match 'everything else' (probably not needed)
/\1/ : and replace with the first backslash-bracketed expression you found
I think this will work (need to recheck my reference):
awk '/crop=([0-9:]*?)/\1/'
yet another way with bash pattern substitution
PAT="pos:665181533 pts:11360 t:11.360000 crop=720:568:0:4 some more words"
RES=${PAT#*crop=}
echo ${RES%% *}
first remove all up to and including crop= found from left to right (#)
then remove all from and including the first space found from right to left (%%)
I am attempting to parse (with sed) just First Last from the following DN(s) returned by the DSCL command in OSX terminal bash environment...
CN=First Last,OU=PCS,OU=guests,DC=domain,DC=edu
I have tried multiple regexs from this site and others with questions very close to what I wanted... mainly this question... I have tried following the advice to the best of my ability (I don't necessarily consider myself a newbie...but definitely a newbie to regex..)
DSCL returns a list of DNs, and I would like to only have First Last printed to a text file. I have attempted using sed, but I can't seem to get the correct function. I am open to other commands to parse the output. Every line begins with CN= and then there is a comma between Last and OU=.
Thank you very much for your help!
I think all of the regular expression answers provided so far are buggy, insofar as they do not properly handle quoted ',' characters in the common name. For example, consider a distinguishedName like:
CN=Doe\, John,CN=Users,DC=example,DC=local
Better to use a real library able to parse the components of a distinguishedName. If you're looking for something quick on the command line, try piping your DN to a command like this:
echo "CN=Doe\, John,CN=Users,DC=activedir,DC=local" | python -c 'import ldap; import sys; print ldap.dn.explode_dn(sys.stdin.read().strip(), notypes=1)[0]'
(depends on having the python-ldap library installed). You could cook up something similar with PHP's built-in ldap_explode_dn() function.
Two cut commands is probably the simplest (although not necessarily the best):
DSCL | cut -d, -f1 | cut -d= -f2
First, split the output from DSCL on commas and print the first field ("CN=First Last"); then split that on equal signs and print the second field.
Using sed:
sed 's/^CN=\([^,]*\).*/\1/' input_file
^ matches start of line
CN= literal string match
\([^,]*\) everything until a comma
.* rest
http://www.gnu.org/software/gawk/manual/gawk.html#Field-Separators
awk -v RS=',' -v FS='=' '$1=="CN"{print $2}' foo.txt
I like awk too, so I print the substring from the fourth char:
DSCL | awk '{FS=","}; {print substr($1,4)}' > filterednames.txt
This regex will parse a distinguished name, giving name and val a capture groups for each match.
When DN strings contain commas, they are meant to be quoted - this regex correctly handles both quoted and unquotes strings, and also handles escaped quotes in quoted strings:
(?:^|,\s?)(?:(?<name>[A-Z]+)=(?<val>"(?:[^"]|"")+"|[^,]+))+
Here is is nicely formatted:
(?:^|,\s?)
(?:
(?<name>[A-Z]+)=
(?<val>"(?:[^"]|"")+"|[^,]+)
)+
Here's a link so you can see it in action:
https://regex101.com/r/zfZX3f/2
If you want a regex to get only the CN, then this adapted version will do it:
(?:^|,\s?)(?:CN=(?<val>"(?:[^"]|"")+"|[^,]+))