I am searching an event field in a file but is giving wrong output. I am searching gpio-keys event in input devices for which I have written a script, but I'm unable to print anything in output file (in my case I am writing in a button device file it is null always). Please help me to figure out this. Where am I doing wrong in script file?
Bash script:
#!/bin/bash
if grep -q "gpio-keys" /proc/bus/input/devices ; then
EVENT=$(cat /proc/bus/input/devices | grep "Handlers=kbd")
foo= `echo $EVENT | awk '{for(i=1;i<=NF;i++) if($i=="evbug")printf($(i-1))}'`
#foo=${EVENT:(-7)}
echo -n $foo > /home/ubuntu/Setups/buttonDevice
fi
i am still not able to get anything in buttondevce
That's no wonder, since in the input line
H: Handlers=kbd event0
there's nowhere the evbug your awk script is looking for.
I my case it is event0 but it may vary also depends on how kernel allows.
If it is event0 or similar, then it's nonsensical to look for evbug. Change the statement
if($i=="evbug")printf($(i-1))
to
if ($i~"event") print $i
(using regular expression match).
I have rewritten my script like above. but through it, I have got two events(event0, event3) but … my input devices are many but i want the gpio-keys event
Aha - in order to take only the handler line from the gpio-keys section, you can use sed with an address range:
EVENT=`sed -n '/gpio-keys/,/Handlers=kbd/s/.*Handlers=kbd //p' </proc/bus/input/devices`
Prakash, I don't have access to your google drive. But I just want to give you some suggestion:-
foo= `echo $EVENT | awk '{for(i=1;i<=NF;i++) if($i=="evbug")printf($(i-1))}'`
This is old style now. Better use like below:-
foo=$(echo $EVENT | awk '{for(i=1;i<=NF;i++) if($i=="evbug")printf($(i-1))}')
Also always use double quotes "" when echoing a variable. See below:-
echo -n "$foo" > /home/ubuntu/Setups/buttonDevice
Try with the below code it will work for you
#!/bin/bash
if grep "gpio-keys" /proc/bus/input/devices >/dev/null ; then
cat /proc/bus/input/devices | grep "Handlers=kbd" | awk '{for(i=1;i<=NF;i++){ if($i ~ /eve/){printf "%s \n", $i} } }') > /home/ubuntu/Setups/buttonDevice
fi
The output in buttonDevice would be
event0
event1
.
.
.
.
event100
Related
Im writing a script that executes dig command on 2 domains, and after next cmd is host on output.
And always i will get for exmaple:
findUserServer=for r in $(dig +short $login.example.COM && dig +short $login.example.ORG); do host $r|awk '{print $NF}';done | awk -F "." '{print $1}';
1 output: >> asdf02 example
asdf02 - its a server name, its always same name starts "asdf".
Question: Have you any idea how to save to the variable only asdf02?
question+: asdf02 woudln't be always first, could be example asdf02
Should i do maybe a sed which looks on 4 first characters? If it's "asdf", then: [...]
Try not to pipe awk commands into each other and so:
for r in $(dig +short $login.example.COM && dig +short $login.example.ORG); do host $r;done | awk -F [.\ ] '/asdf02/ { print $10 }'
We use both a space and . as delimiters and then pattern match the output for the occurance of asdf02. If we find is, we print the address.
Run that through shellcheck.net ...
Try this.
findUserServer="$( for end in COM ORG; do
host $( dig +short $login.example.$end );
done | sed -n '/ asdf/{ s/^.* //; s/[.].*//; p; }' )"
This will run 2 digs and pipe the collective output through sed,
which will ignore lines that don't have asdf, and strip the matches clean for you.
Let me know if I missed details, because I don't have those exact values available.
I have a file which is created after some manual configuration.
I need to check this file automatically with a shell script.
The file looks like this:
eth0;eth0;1c:98:ec:2a:1a:4c
eth1;eth1;1c:98:ec:2a:1a:4d
eth2;eth2;1c:98:ec:2a:1a:4e
eth3;eth3;1c:98:ec:2a:1a:4f
eth4;eth4;48:df:37:58:da:44
eth5;eth5;48:df:37:58:da:45
eth6;eth6;48:df:37:58:da:46
eth7;eth7;48:df:37:58:da:47
I want to compare it to a pattern like this:
eth0;eth0;*
eth1;eth1;*
eth2;eth2;*
eth3;eth3;*
eth4;eth4;*
eth5;eth5;*
eth6;eth6;*
eth7;eth7;*
If I would only have to check this pattern I could run this loop:
c=0
while [ $c -le 7 ]
do
if [ "$(grep "eth"${c}";eth"${c}";*" current_mapping)" ];
then
echo "eth$c ok"
fi
(( c++ ))
done
There are 6 or more different patterns possible. A pattern could also look like this for example (depending and specific configuration requests):
eth4;eth0;*
eth5;eth1;*
eth6;eth2;*
eth7;eth3;*
eth0;eth4;*
eth1;eth5;*
eth2;eth6;*
eth3;eth7;*
So I don't think I can run a standard grep per line command in a loop. The eth numbers are not consistently the same.
Is it possible somehow to compare the whole file to pattern like it would be possible with grep for a single line?
Assuming file is your data file and patt is your file that contains above pattern. You can use this grep -f in conjunction with sed in a process substitution that replaces * with .* and ? with . to make it a workable regex.
grep -f <(sed 's/\*/.*/g; s/?/./g' patt) file
eth0;eth0;1c:98:ec:2a:1a:4c
eth1;eth1;1c:98:ec:2a:1a:4d
eth2;eth2;1c:98:ec:2a:1a:4e
eth3;eth3;1c:98:ec:2a:1a:4f
eth4;eth4;48:df:37:58:da:44
eth5;eth5;48:df:37:58:da:45
eth6;eth6;48:df:37:58:da:46
eth7;eth7;48:df:37:58:da:47
I wrote this loop now and it does the job (current_mapping being the file with the content in the first code block of the question). I would have to create arrays with different patterns and use a case for every pattern. I was just wondering if there is something like grep for multiple lines, that could the same without writing this loop.
array=("eth0;eth0;*" "eth1;eth1;*" "eth2;eth2;*" "eth3;eth3;*" "eth4;eth4;*" "eth5;eth5;*" "eth6;eth6;*" "eth7;eth7;*")
c=1
while [ $c -le 8 ]
do
if [ ! "$(sed -n "${c}"p current_mapping | grep "${array[$c-1]}")" ];
then
echo "somethings wrong"
fi
(( c++ ))
done
Try any:
grep -P '(eth[0-9]);\1'
grep -E '(eth[0-9]);\1'
sed -n '/\(eth[0-9]\);\1/p'
awk -F';' '$1 == $2'
There are commands only. Apply them to a pipe or file.
Updated the answer after the question was edited.
As we can see the task requirements are as follows:
a file (a set of lines) formatted like ethN;ethM;MAC
examine each line for equality ethN and ethM
if they are equal, output a string ethN ok
If I understand the task correctly we can achieve this using the following code without loops:
awk -F';' '$1 == $2 { print $1, "ok" }'
I'm trying to search through files and extract two pieces of relevant information every time they appear in the file. The code I currently have:
#!/bin/bash
echo "Utilized reads from ustacks output" > reads.txt
str1="utilized reads:"
str2="Parsing"
for file in /home/desaixmg/novogene/stacks/sample01/conda_ustacks.o*; do
reads=$(grep $str1 $file | cut -d ':' -f 3
samples=$(grep $str2 $file | cut -d '/' -f 8
echo $samples $reads >> reads.txt
done
It is doing each line for the file (the files have varying numbers of instances of these phrases) and gives me the output per row for each file:
PopA_15.fq 1081264
PopA_16.fq PopA_17.fq 1008416 554791
PopA_18.fq PopA_20.fq PopA_21.fq 604610 531227 595129
...
I want it to match each instance (i.e. 1st instance of both greps next two each other):
PopA_15.fq 1081264
PopA_16.fq 1008416
PopA_17.fq 554791
PopA_18.fq 604610
PopA_20.fq 531227
PopA_21.fq 595129
...
How do I do this? Thank you
Considering that your Input_file is same as sample shown and number of columns are even on each line with 1 PopA value and other will be with digit values. Following awk may help you in same.
awk '{for(i=1;i<=(NF/2);i++){print $i,$((NF/2)+i)}}' Input_file
Output will be as follows.
PopA_15.fq 1081264
PopA_16.fq 1008416
PopA_17.fq 554791
PopA_18.fq 604610
PopA_20.fq 531227
PopA_21.fq 595129
In case you want to pass output of a command to awk command then you could do like your command | awk command... no need to add Input_file to above awk command.
This is what ended up working for me...any tips for more efficient code are definitely welcome
#!/bin/bash
echo "Utilized reads from ustacks output" > reads.txt
str1="utilized reads:"
str2="Parsing"
for file in /home/desaixmg/novogene/stacks/sample01/conda_ustacks.o*; do
reads=$(grep $str1 $file | cut -d ':' -f 3)
samples=$(grep $str2 $file | cut -d '/' -f 8)
paste <(echo "$samples" | column -t) <(echo "$reads" | column -t) >> reads.txt
done
This provides the desired output described above.
This question already has answers here:
How to show only next line after the matched one?
(14 answers)
Closed 6 years ago.
I'm trying to get the current track running from 'cmus-remote -Q'
Its always underneath of this line
tag genre Various
<some track>
Now, I need to keep it simple because I want to add it to my i3 bar. I used
cmus-remote -Q | grep -A 1 "tag genre"
but that grep's the 'tag' line AND the line underneath.
I want ONLY the line underneath.
With sed:
sed -n '/tag genre/{n;p}'
Output:
$ cmus-remote -Q | sed -n '/tag genre/{n;p}'
<some track>
If you want to use grep as the tool for this, you can achieve it by adding another segment to your pipeline:
cmus-remote -Q | grep -A 1 "tag genre" | grep -v "tag genre"
This will fail in cases where the string you're searching for is on two lines in a row. You'll have to define what behaviour you want in that case if we're going to program something sensible for it.
Another possibility would be to use a tool like awk, which allows for greater compexity in the line selection:
cmus-remote -Q | awk '/tag genre/ { getline; print }'
This searches for the string, then gets the next line, then prints it.
Another possibility would be to do this in bash alone:
while read line; do
[[ $line =~ tag\ genre ]] && read line && echo "$line"
done < <(cmus-remote -Q)
This implements the same functionality as the awk script, only using no external tools at all. It's likely slower than the awk script.
You can use awk instead of grep:
awk 'p{print; p=0} /tag genre/{p=1}' file
<some track>
/tag genre/{p=1} - sets a flag p=1 when it encounters tag genre in a line.
p{print; p=0} when p is non-zero then it prints a line and resets p to 0.
I'd suggest using awk:
awk 'seen && seen--; /tag genre/ { seen = 1 }'
when seen is true, print the line.
when seen is true, decrement the value, so it will no longer true after the desired number of lines are printed
when the pattern matches, set seen to the number of lines to be printed
I have written a line that finds and returns the full path to a desired file. The output is as follows:
/home/ke/Desktop/b/o/r/files.txt:am.torrent
/home/ke/Desktop/y/u/n/u/s/files.txt:asd.torrent
I have to modify the output like this:
bor
yunus
How do I do that?
Thanks in advance.
This should work for you:
your_script.sh | sed 's,.*Desktop,,' | sed 's,[^/]*$,,' | sed s,/,,g
or, even better:
your_script.sh | sed 's,.*Desktop,,;s,[^/]*$,,;s,/,,g'
With sed. echo '/home/ke/Desktop/b/o/r/files.txt:am.torrent' | sed -e 's+/++g' -e 's/^.*Desktop//' -e 's/files.txt:.*$//'. This is a fairly trivial solution, and I'm sure there are better ones.
Id resort to awk:
BEGIN { FS="/" }
{
for(i=1;i<NF;i++)
if (length($i) == 1)
a[NR]=a[NR]""$i
}
END {
for (i in a)
print a[i]
}
use it like this:
$ awk -f script.awk input
bor
yunus
or if you have your data in a variable:
$ awk -f script.awk <<< $data
it's not a nice/tidy solution, but bash parameter expansion is a powerful tool. So could not resist providing an example
[]l="/home/ke/Desktop/b/o/r/files.txt:am.torrent"
[]m=${l##*Desktop/}
[]n=${m%%/files.txt*}
[]k=${n//\//}
[]echo $m
b/o/r/files.txt:am.torrent
[]echo $n
b/o/r
[]echo $k
bor
You can see how nicely bash is replacing the variable step by step without using any external program (btw [] is PS1, prompt)
There can be many more ways to do it. I got another one while writing the first
[]l="/home/ke/Desktop/b/o/r/files.txt:am.torrent"
[]m=${l/*Desktop\//}
[]n=${m/\/files.txt*/}
[]k=${n//\//}
[]echo $m
b/o/r/files.txt:am.torrent
[]echo $n
b/o/r
[]echo $k
bor
Try some more,