Using grep regex to select to first hyphen - regex

echo "Linux/DEB/mainbinary-0.1.20190424165331-0-armdef.deb" | grep -oE "([^\/]+$)"
This prints just the filename, without the directory structure, but I cannot manage to print just mainbinary from that string. Suggestions?


And a sed alternative to PS.'s great grep -oP
echo "Linux/DEB/mainbinary-0.1.20190424165331-0-armdef.deb" |sed -r 's#^.*/([^-]+).*#\1#'
mainbinary


echo "Linux/DEB/mainbinary-0.1.20190424165331-0-armdef.deb" |grep -oP '.*/\K[^-]+'
mainbinary
This will scan till last / and ignore everything to its left and keep moving until - (excluding)


With any awk in any shell on any UNIX machine:
$ echo "Linux/DEB/mainbinary-0.1.20190424165331-0-armdef.deb" | awk -F'[/-]' '{print $3}'
mainbinary

Related

How to use sed to identify a string in brackets?

I want to find the string in that is placed with in the brackets. How do I use sed to pull the string?
# cat /sys/block/sdb/queue/scheduler
noop anticipatory deadline [cfq]
I'm not getting the exact result
# cat /sys/block/sdb/queue/scheduler | sed 's/\[*\]//'
noop anticipatory deadline [cfq
I'm expecting an output
cfq

It can be easier with grep, if it happens to be changing the position in which the text in between brackets is located:
$ grep -Po '(?<=\[)[^]]*' file
cfq
This is look-behind: whenever you find a string [, start fetching all the characters up to a ].
See another example:
$ cat a
noop anticipatory deadline [cfq]
hello this [is something] we want to [enclose] yeah
$ grep -Po '(?<=\[)[^]]*' a
cfq
is something
enclose
You can also use awk for this, in case it is always in the same position:
$ awk -F[][] '{print $2}' file
cfq
It is setting the field separators as [ and ]. And from that, prints the second one.
And with sed:
$ sed 's/[^[]*\[\([^]]*\).*/\1/g' file
cfq
It is a bit messy, but basically it is looking from the block of text in between [] and prints it back.

I found one possible solution-
cut -d "[" -f2 | cut -d "]" -f1
so the exact solution is
# cat /sys/block/sdb/queue/scheduler | cut -d "[" -f2 | cut -d "]" -f1

Another potential solution is awk:
s='noop anticipatory deadline [cfq]'
awk -F'[][]' '{print $2}' <<< "$s"
cfq

Another way by gnu grep :
grep -Po "\[\K[^]]*" file
with pure shell:
while read line; do [[ "$line" =~ \[([^]]*)\] ]] && echo "${BASH_REMATCH[1]}"; done < file

Another awk
echo 'noop anticipatory deadline [cfq]' | awk '{gsub(/.*\[|\].*/,x)}8'
cfq

perl -lne 'print $1 if(/\[([^\]]*)\]/)'
Tested here

(GNU)Sed: how to replace any character from nth character to nth+10?

I need to replace characters from 10th to 20th in the string which looks like that:
123456789012345678901234567890
So far I've tried:
a)
Works for the 10th character ONLY:
echo "123456789012345678901234567890" | sed 's/./X/10'
b)
Doesn't work on the range:
echo "123456789012345678901234567890" | sed 's/./X/10,20'
echo "123456789012345678901234567890" | sed 's/./X/10\,20'
echo "123456789012345678901234567890" | sed 's/./X/\{10,20\}'
echo "123456789012345678901234567890" | sed 's/./X/\{10\,20\}'
Does not work and I get error
unknown option to `s'
So - the question is - how do I make this to work:
echo "123456789012345678901234567890" | sed 's/./X/10,20'

Try:
$ sed -r "s/^(.{9})(.{11})/\1XXXXXXXXXX/" <<< 123456789012345678901234567890
123456789XXXXXXXXXX1234567890

It is a complex sed problem, I could just find this solution:
$ sed 's/^\(.\{10\}\)\(.\{10\}\)/\1XXXXXXXXXX/' <<< 123456789012345678901234567890
1234567890XXXXXXXXXX1234567890
With awk it looks nicer:
$ awk 'BEGIN{FS=OFS=""} {for (i=10;i<=20;i++) $i="X"} {print}' <<< 123456789012345678901234567890
123456789XXXXXXXXXXX1234567890

You can do it with bash parameter substitution like this:
#!/bin/bash
s="123456789012345678901234567890"
l=${s:0:9} # Extract left part
m=${s:10:11} # Extract middle part
r=${s:20} # Extract right part
# Diddle with middle part to your heart's content and re-assemble "$l$m$r" when done
m=$(sed 's/./X/g' <<<$m)
See here for more explanation and examples.
Or, you can do this:
transform the row of letters into a column so each is on its own line
apply your edits to LINES 10 through 20 (as opposed to characters 10 through 20)
transform column of letters back into a row (by deleting linefeeds)
as shown in the one-liner below:
$ echo "123456789012345678901234567890" | sed "s/\(.\)/\1\n/g" | sed "10,20s/./X/" | tr -d "\n"

I know, that it looks ugly, but:
echo "123456789012345678901234567890" | \
sed 's/^\(.\{10\}\).\{10\}\(.*\)/\1XXXXXXXXXX\2/'

Without placing multiple X in sed command:
sed -r 's/^(.{9})(.{10,20})(.*)$/\1\n\2\n\3/' | sed -e '2s/./X/g' -e 'N;N;s/\n//g'

To replace the 10th to 20th characters, inclusive, try:
echo 123456789012345678901234567890 | sed 's/\(.\{9\}\).\{11\}/\1XXXXXXXXXX/'
123456789XXXXXXXXXX1234567890
With the GNU sed, you can use the -r switch to remove most of the backslashes:
echo 123456789012345678901234567890 | sed -r 's/(.{9}).{11}/\1XXXXXXXXXX/'
Or the naive approach also works here:
echo 123456789012345678901234567890 | sed 's/\(.........\).........../\1XXXXXXXXXX/'

This might work for you (GNU sed):
sed ':a;/.\{9\}X\{11\}/!s/\(.\{9\}X*\)./\1X/;ta' file
or with a bit of syntactic sugar:
sed -r ':a;/.{9}X{11}/!s/(.{9}X*)./\1X/;ta' file

How to extract a number from a string using grep and regex

I make a cat of a file and apply on it a grep with a regular expression like this
cat /tmp/tmp_file | grep "toto.titi\[[0-9]\+\].tata=55"
the command display the following output
toto.titi[12].tata=55
is it possible to modify my grep command in order to extract the number 12 as displayed output of the command?

You can grab this in pure BASH using its regex capabilities:
s='toto.titi[12].tata=55'
[[ "$s" =~ ^toto.titi\[([0-9]+)\]\.tata=[0-9]+$ ]] && echo "${BASH_REMATCH[1]}"
12
You can also use sed:
sed 's/toto.titi\[\([0-9]*\)\].tata=55/\1/' <<< "$s"
12
OR using awk:
awk -F '[\\[\\]]' '{print $2}' <<<"$s"
12

use lookahead
echo toto.titi[12].tata=55|grep -oP '(?<=\[)\d+'
12
without perl regex,use sed to replace "["
echo toto.titi[12].tata=55|grep -o "\[[0-9]\+"|sed 's/\[//g'
12

Pipe it to sed and use a back reference:
cat /tmp/tmp_file | grep "toto.titi\[[0-9]\+\].tata=55" | sed 's/.*\[(\d*)\].*/\1/'

sed or awk to capture part of url

I am not very experienced with regular expressions and sed/awk scripting.
I have urls that are similar to the following torrent url:
http://torcache.net/torrent/D7249CD9AF321C8578B3A7007ABBDD63B0475EEB.torrent?title=[kickass.to]against.the.ropes.by.carly.fall.epub.torrent
I would like to have sed or awk script extract the text after the title i.e
from the example above just get:
[kickass.to]against.the.ropes.by.carly.fall.epub.torrent

A simple approach with awk: use the = as the field separator:
awk -F"=" '{print $2}'
Thus:
echo "http://torcache.net/torrent/D7249CD9AF321C8578B3A7007ABBDD63B0475EEB.torrent?title=[kickass.to]against.the.ropes.by.carly.fall.epub.torrent" | awk -F"=" '{print $2}'
[kickass.to]against.the.ropes.by.carly.fall.epub.torrent

Just remove everything before the title=: sed 's/.*title=//'
$ echo "http://torcache.net/torrent/D7249CD9AF321C8578B3A7007ABBDD63B0475EEB.torrent?title=[kickass.to]against.the.ropes.by.carly.fall.epub.torrent" | sed 's/.*title=//'
[kickass.to]against.the.ropes.by.carly.fall.epub.torrent

Let's say:
s='http://torcache.net/torrent/D7249CD9AF321C8578B3A7007ABBDD63B0475EEB.torrent?title=[kickass.to]against.the.ropes.by.carly.fall.epub.torrent'
Pure BASH solution:
echo "${s/*title=}"
[kickass.to]against.the.ropes.by.carly.fall.epub.torrent
OR using grep -P:
echo "$s"|grep -oP 'title=\K.*'
[kickass.to]against.the.ropes.by.carly.fall.epub.torrent

By using sed (no need to mention title in the regexp in your example) :
sed 's/.*=//'
An another solution exists with cut, another standard unix tool :
cut -d= -f2

Can not extract the capture group with either sed or grep

I want to extract the value pair from a key-value pair syntax but I can not.
Example I tried:
echo employee_id=1234 | sed 's/employee_id=\([0-9]+\)/\1/g'
But this gives employee_id=1234 and not 1234 which is actually the capture group.
What am I doing wrong here? I also tried:
echo employee_id=1234| egrep -o employee_id=([0-9]+)
but no success.

1. Use grep -Eo: (as egrep is deprecated)
echo 'employee_id=1234' | grep -Eo '[0-9]+'
1234
2. using grep -oP (PCRE):
echo 'employee_id=1234' | grep -oP 'employee_id=\K([0-9]+)'
1234
3. Using sed:
echo 'employee_id=1234' | sed 's/^.*employee_id=\([0-9][0-9]*\).*$/\1/'
1234

To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:
$ regex="$precedes_regex\K($capture_regex)(?=$follows_regex)"
$ echo $some_string | grep -oP "$regex"
so
# matches and returns b
$ echo "abc" | grep -oP "a\K(b)(?=c)"
b
# no match
$ echo "abc" | grep -oP "z\K(b)(?=c)"
# no match
$ echo "abc" | grep -oP "a\K(b)(?=d)"

Using awk
echo 'employee_id=1234' | awk -F= '{print $2}'
1234

use sed -E for extended regex
echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/\1/g'

You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:
foo='employee_id=1234'
var=${foo%%=*}
value=${foo#*=}
 
$ echo "var=${var} value=${value}"
var=employee_id value=1234