I have a script that outputs file paths (via find) :
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3-something16
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something1
/foo/bar/example/foo/bar/example/foo-bar-example/1.2-something5
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something2
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something10
How can I list them in Bash so that they are in ascending numeric order based on the number at the end and regardless of the version ( 1.2, 1.2.3 or 1.2.3.4)
Ps : the something part can eventually contain a dash.
Desired output :
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something1
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something2
/foo/bar/example/foo/bar/example/foo-bar-example/1.2-something5
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something10
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3-something16
I used a temporary sentinel character to delimit the number at the end but it's a little bit complicated in my case.
Extract the number at the end, prepend it to all the lines, sort numerically and finally delete that number:
sed -r 's/(.*)([^0-9])([0-9]+)$/\3\1\2\3/' file | sort -n | sed 's/^[0-9]*//'
With your input, this returns:
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something1
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something2
/foo/bar/example/foo/bar/example/foo-bar-example/1.2-something5
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something10
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3-something16
I think this is the Schwartzian transform.
By pieces:
$ sed -r 's/(.*)([^0-9])([0-9]+)$/\3\1\2\3/' a
16/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3-something16
1/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something1
5/foo/bar/example/foo/bar/example/foo-bar-example/1.2-something5
2/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something2
10/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something10
#^
#note the numbers here
$ sed -r 's/(.*)([^0-9])([0-9]+)$/\3\1\2\3/' a | sort -n
1/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something1
2/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something2
5/foo/bar/example/foo/bar/example/foo-bar-example/1.2-something5
10/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something10
16/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3-something16
# ^
# now it is ordered
$ sed -r 's/(.*)([^0-9])([0-9]+)$/\3\1\2\3/' a | sort -n | sed 's/^[0-9]*//'
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something1
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something2
/foo/bar/example/foo/bar/example/foo-bar-example/1.2-something5
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something10
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3-something16
You can use this sed + sort + awk:
sed -E 's/.*[^0-9]([0-9]+)$/\1 &/' file | sort -n | awk '{print $2}'
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something1
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something2
/foo/bar/example/foo/bar/example/foo-bar-example/1.2-something5
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3.4-something10
/foo/bar/example/foo/bar/example/foo-bar-example/1.2.3-something16
If you can ensure that you will never have a '#' character in your file, you can try:
sed -e 's/something/#/g' filename.txt | sort -t# -k2 -n | sed -e 's/#/something/g'
Related
(Need in bash linux)I have a file with numbers like this
1.415949602
91.09582241
91.12042924
91.40270349
91.45625033
91.70150341
91.70174342
91.70660043
91.70966213
91.72597066
91.7287678315
91.7398645966
91.7542977976
91.7678146465
91.77196659
91.77299733
abcdefghij
91.7827827
91.78288651
91.7838959
91.7855
91.79080605
91.80103075
91.8050505
sed 's/^91\.//' file (working)
Any way possible I can do these 3 steps?
1st I try this
cat input | tr -d 91. > 1.txt (didnt work)
cat input | tr -d "91." > 1.txt (didnt work)
cat input | tr -d '91.' > 1.txt (didnt work)
then
grep -x '.\{10\}' (working)
then
grep "^[6-9]" (working)
Final 1 line solution
cat input.txt | sed 's/\91.//g' | grep -x '.\{10\}' | grep "^[6-9]" > output.txt
Your "final" solution:
cat input.txt |
sed 's/\91.//g' |
grep -x '.\{10\}' |
grep "^[6-9]" > output.txt
should avoid the useless cat, and also move the backslash in the sed script to the correct place (and I added a ^ anchor and removed the g flag since you don't expect more than one match on a line anyway);
sed 's/^91\.//' input.txt |
grep -x '.\{10\}' |
grep "^[6-9]" > output.txt
You might also be able to get rid of at least one useless grep but at this point, I would switch to Awk:
awk '{ sub(/^91\./, "") } /^[6-9].{9}$/' input.txt >output.txt
The sub() does what your sed replacement did; the final condition says to print lines which match the regex.
The same can conveniently, but less readably, be written in sed:
sed -n 's/^91\.([6-9][0-9]\{9\}\)$/\1/p' input.txt >output.txt
assuming your sed dialect supports BRE regex with repetitions like [0-9]\{9\}.
I am trying to grab some content, but there are multiple instances of it in the same line. I am using this command.
grep -o -m 1 -P '(?<=sk).*(?=fa)' test.txt | head -1
However, the search ends after the second/last match. Running it on Ubuntu 14.04.2
test.txt: skjahfasdkl aklsdj laks skjahfasdkl aklsdj laks
Current Output: jahfasdkl aklsdj laks skjah
Desired output: jah
You just need non-greedy:
grep -m1 -oP '(?<=sk).*?(?=fa)' file | head -1
# ...................^^^
The -m1 will stop after the first line, but you still need head to limit to the first match.
It's greedy match, you want to treat space as delimiters so specify the match to nonspace chars, i.e.
... '(?<=sk)[^ ]*(?=fa)'
if the condition is non spaces between sk and fa ( matching in words ), you can use can use [^ ]* instead .*, as the following:
grep -o -m 1 -P '(?<=sk)[^ ]*(?=fa)' test.txt | head -1
else you can use this :
sed -e "s/sk\(.*\)fa.*$/\1/g" test.txt | sed -e "s/fa.*$//g"
test :
echo "skjahz z zfasdkl aklsdj laks skjahppppfasdkl aklsdj laks" | sed -e "s/sk\(.*\)fa.*$/\1/g" | sed -e "s/fa.*$//g"
#jahz z z
If you consider non-grep answer then this gnu-awk can do the job:
awk -v FPAT='sk[^[:blank:]]*fa' '{gsub(/^sk|fa$/, "", $1); print $1; exit}' file
I have some file xxx.conf in text format. I have some text "disablelog = 1" in this file.
When I use
grep -r "disablelog" oscam.conf
output is
disablelog = 1
But i need only value 1.
Do you have some idea please?
one way is to use awk to print just the value
grep -r "disablelog" oscam.conf | awk '{print $3}'
you could also use sed to replace diablelog = with empty
grep -r 'disablelog' oscam.conf | sed -e 's/disablelog = //'
If you also want to get the lines with or without space before and after = use
grep -r 'disablelog' oscam.conf | sed 's/disablelog\s*=\s*//'
above command will also match
disablelog=1
Assuming you need it as a var in a script:
#!/bin/bash
DISABLELOG=$(awk -F= '/^.*disablelog/{gsub(/ /,"",$2);print $2}' /path/to/oscam.conf)
echo $DISABLELOG
When calling this script, the output should be 1.
Edit: No matter wether there is whitespace or not between the equals sign and the value, the above will handle that. The regex should be anchored in either way to improve performance.
Try:
grep -r "disablelog" oscam.conf | awk -F= '{print $2}'
Just for fun a solution without awk
grep -r disablelog | cut -d= -f2 | xargs
xargs is used here to trim the whitespace
I have something about 100 files with the following syntax
ahfsdjfhdfhj_EPI_34_fdsafasdf
asdfasdf_EPI_2_fdsf
hfdjh_EPI_8_dhfffffffffff
ffffffffffasdfsdf_EPI_1_fyyy44
...
There is always EPI_NUMBER. How can I sort it by this number?
From your example it appears that delimiter is _ and text EPI_nnn comes at the same position after delimiter _. If that is always the case then you can use following command to sort the file:
sort -n -t "_" -k 3 file.txt
UPDATE:
If position of EPI_ text is not fixed then use following shell command:
sed 's/^\(.*EPI_\)\(.*\)$/\2##\1/' file.txt | sort -n -t "_" -k1 | sed 's/^\(.*\)##\(.*\)$/\2\1/'
If Perl is okay you can:
print sort foo <>;
sub foo {
($x = $a) =~s/.*EPI_(\d+).*/$1/;
($y = $b) =~s/.*EPI_(\d+).*/$1/;
return $x <=> $y;
}
and use it as:
perl prg.pl inputfile
See it
sed -e 's/EPI_/EPI /' file1 file2 ...|sort -n -k 2 -t ' '
Pipe that to sed -e 's/ /_/' to get back the original form.
This might work for you:
ls | sed 's/.*EPI_\([0-9]*\)/\1 &/' | sort -n | sed 's/\S* //'
I use the df command in a bash script:
df . -B MB | tail -1 | awk {'print $4'} | grep .[0-9]*
This script returns:
99%
But I need only numbers (to make the next comparison).
If I use the grep regex without the dot:
df . -B MB | tail -1 | awk {'print $4'} | grep .[0-9]*
I receive nothing.
How to fix?
If you try:
echo "99%" |grep -o '[0-9]*'
It returns:
99
Here's the details on the -o (or --only-matching flag) works from the grep manual page.
Print only the matched (non-empty) parts of matching lines, with each such part on a separate output line. Output lines use the same delimiters as input, and delimiters are null bytes if -z (--null-data) is also used (see Other Options).
grep will print any lines matching the pattern you provide. If you only want to print the part of the line that matches the pattern, you can pass the -o option:
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Like this:
echo 'Here is a line mentioning 99% somewhere' | grep -o '[0-9]+'
How about:
df . -B MB | tail -1 | awk {'print $4'} | cut -d'%' -f1
No need to used grep here, Try this:
df . -B MB | tail -1 | awk {'print substr($5, 1, length($5)-1)'}
function getPercentUsed() {
$sys = system("df -h /dev/sda6 --output=pcent | grep -o '[0-9]*'", $val);
return $val[0];
}
Don't use more commands than necessary, leave away tail, grep and cut. You can do this with only (a simple) awk
PS: giving a block-size en print only de persentage is a bit silly ;-) So leave also away the "-B MB"
df . |awk -F'[multiple field seperators]' '$NF=="Last field must be exactly --> mounted patition" {print $(NF-number from last field)}'
in your case, use:
df . |awk -F'[ %]' '$NF=="/" {print $(NF-2)}'
output: 81
If you want to show the percent symbol, you can leave the -F'[ %]' away and your print field will move 1 field further back
df . |awk '$NF=="/" {print $(NF-1)}'
output: 81%
You can use Perl style regular expressions as well. A digit is just \d then.
grep -Po "\\d+" filename
-P Interpret PATTERNS as Perl-compatible regular expressions (PCREs).
-o Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.