I am trying to replace part of the string, but can not find a proper regex for sed to execute it properly.
I have a string
/abc/foo/../bar
And I would like to achive the following result:
/abc/bar
I have tried to do it using this command:
echo $string | sed 's/\/[^:-]*\..\//\//'
But as result I am getting just /bar.
I understand that I must use group, but I just do not get it. Could you, please, help me to find out this group that could be used?
You can use
#!/bin/bash
string='/abc/foo/../bar'
sed -nE 's~^(/[^/]*)(/.*)?/\.\.(/[^/]*).*~\1\3~p' <<< "$string"
See the online demo. Details:
-n - suppresses default line outputE - enables POSIX ERE regex syntax^ - start of string(/[^/]*) - Group 1: a / and then zero or more chars other than /(/.*)? - an optional group 2: a / and then any text/\.\. - a /.. fixed string(/[^/]*) - Group 3: a / and then zero or more chars other than /.* - the rest of the string.\1\3 replaces the match with Group 1 and 3 values concatenatedp only prints the result of successful substitution.You can use a capture group for the first part and then match until the last / to remove.
As you are using / to match in the pattern, you can opt for a different delimiter.
#!/bin/bash
string="/abc/foo/../bar"
sed 's~\(/[^/]*/\)[^:-]*/~\1~' <<< "$string"
The pattern in parts:
\( Capture group 1
/[^/]*/ Match from the first till the second / with any char other than / in between\) Close group 1[^:-]*/ Match optional chars other than : and - then match /Output
/abc/bar
Using sed
$ sed 's#^\(/[^/]*\)/.*\(/\)#\1\2#' input_file
/abc/bar
or
$ sed 's#[^/]*/[^/]*/##2' input_file
/abc/bar
Using awk
string='/abc/foo/../bar'
awk -F/ '{print "/"$2"/"$NF}' <<< "$string"
#or
awk -F/ 'BEGIN{OFS=FS}{print $1,$2,$NF}' <<< "$string"
/abc/bar
Using bash
string='/abc/foo/../bar'
echo "${string%%/${string#*/*/}}/${string##*/}"
/abc/bar