linux shell storing path - linux

I want to find the path of folder and store it into a variable
#!/bin/bash
howdy=$(whereis yum.repos.d)
howdy=$howdy"/remi.repo"
echo $howdy
The issue is that when I want to use the variable $howdy , it will output
yum.repos: /etc/yum.repos.d/remi.repo
I want it to be just the path
/etc/yum.repos.d/remi.repo
so I could use it in my code

You can use Parameter Expansion:
${howdy#yum.repos: }

You can erase the first part of the string:
echo ${howdy##*: }
Or you can pipe it through cut -d' ' -f2

you can use awk to format the output:
howdy=$(whereis yum.repos.d| awk '{print $2}')
Full code:
#!/bin/bash
howdy=$(whereis yum.repos.d| awk '{print $2}')
howdy=$howdy"/remi.repo"
echo $howdy
this will output: /etc/yum.repos.d/remi.repo

How about:
howdy=$(find /etc -type f -name remi.repo)

whereis yum.repos.d | awk -F: '{print $2}' will give strip off the first part

Related

Sum out of grep -c

I am trying to find the number an even occured in my log file.
Command:
grep -Eo "2016-08-30" applciationLog.log* -c
Output:
applciationLog.log.1:0
applciationLog.log.2:0
applciationLog.log.3:0
applciationLog.log.4:0
applciationLog.log.5:7684
applciationLog.log.6:9142
applciationLog.log.7:8699
applciationLog.log.8:0
What I actually need is sum of all these values 7684 + 9142 + 8699 = 25525. Any suggestion I can do it? Anything I can append to the grep to enable it.
Any help or pointers are welcome and appreciated.
If you want to keep your grep command, pipe its output to awk, the quick and dirty way is down here:
grep -Eo "aaa" -c aaa.txt bbb.txt -c | awk 'BEGIN {cnt=0;FS=":"}; {cnt+=$2;}; END {print cnt;}'
Or use use awk regex directly:
awk 'BEGIN {cnt=0}; {if(/aaa/) {cnt+=1;}}; END {print cnt;}' aaa.txt bbb.txt
As addition to the already given answer by ghoti:
You can avoid awk -F: by using grep -h:
grep -c -h -F "2016-08-30" applicationLog.log* | awk '{n+=$0} END {print n}'
This means no filenames and only the counts are printed by grep and we can use the first field for the addition in awk.
See if this works for you:
grep -Eo "2016-08-30" applciationLog.log* -c | awk -F':' 'BEGIN {sum = 0;} {sum += $2;} END {print sum;}'
We use awk to split each line up with a delimeter of :, sum up the numbers for each line, and print the result at the end.
The grep command doesn't do arithmetic, it just finds lines that match regular expressions.
To count the output you already have, I'd use awk.
grep -c -F "2016-08-30" applciationLog.log* | awk -F: '{n+=$2} END {print n}'
Note that your grep options didn't make sense -- -E tells the command to use Extended regular expressions, but you're just looking for a fixed string (the date). So I swapped in the -F option instead. And -o tells grep to print the matched text, which you've overridden with -c, so I dropped it.
An alternative using for-loop and arithmetic expansion could be:
x=0
for i in $(grep -hc "2016-08-30" applciationLog.log*);do
x=$((x+i))
done
echo "$x"
An easy alternative is to merge all the files before grep sees them:
cat applciationLog.log* | grep -Eo "2016-08-30" -c
In my directory have have hundreds of files, each file contains lot of text along with a lines similar to this-
Job_1-Run.log:[08/27/20 01:28:40] Total Jobs Cancelled for Job_1_set0 = 10
I do
grep '^Total Jobs Cancelled' ./*
to get that above line.
Then I do a pipe
| awk 'BEGIN {cnt=0;FS="="}; {cnt+=$2;}; END {print cnt;}'
so my final command is-
grep '^Total Jobs Cancelled' ./* | awk 'BEGIN {cnt=0;FS="="}; {cnt+=$2;};END {print cnt;}'
and result is the sum. e.g. -
900
I am using Cmder # https://cmder.net/
Thanks to the answer by #alagner, #john above

can not use unix $variable in awk command [duplicate]

This question already has answers here:
Using awk with variables
(3 answers)
Closed 8 years ago.
I have following variable set in my unix environment. If i try to use it in awk command its not working but the same command is working when i dont use $b variable
$b="NEW"
when i try following command it is not working
echo "$a" | tr [a-z] [A-Z] |awk -v RS=, '/TABLE/&&/CREATE/&&/`echo ${b}`/{print $NF}'
But, if i replace the $b value to NEW as below its working
echo "$a" | tr [a-z] [A-Z] |awk -v RS=, '/TABLE/&&/CREATE/&&/NEW/{print $NF}'
You cannot use a bash var inside awk like that. Instead, use:
echo "$a" | tr [a-z] [A-Z] | awk -v RS=, -v myvar=$b '/TABLE/&&/CREATE/&& $0~myvar {print $NF}'
See an example:
$ var="hello"
$ awk -v text=$var 'BEGIN{print text}'
hello
Also, to me it works with tr 'a-z' 'A-Z' instead of tr [a-z] [A-Z]. And based on Mark Setchell suggestion, you can skip it by using the IGNORECASE = 1:
echo "$a" | awk -v RS=, -v myvar=$b 'BEGIN{IGNORECASE=1} /TABLE/&&/CREATE/&& $0~myvar {print $NF}'
Regarding your question:
if i replace the $b value to NEW as below its working
It works because the value of your variable is NEW and what you end up doing is using that in the regex, which is exactly how it is supposed to be done.
about your second question:
can not use unix $variable in awk command
You cannot use shell variables in awk like that. You need to create an awk variable by using -v option and assigning your bash variable.
awk -v awkvar="$bashvar" '/ /{ ... }'
This makes your existing syntax as:
echo "$a" | tr [a-z] [A-Z] | awk -v RS=, -v var="$b" '/TABLE/&&/CREATE/&&/var/{print $NF}'
This again won't work because inside /../ variables are not interpolated, meaning they are considered literally. So, you need to do:
echo "$a" | tr [a-z] [A-Z] |awk -v RS=, -v var="$b" '/TABLE/&&/CREATE/&&$0~var{print $NF}'

Bash script to remove last three charater in a file name

For ex the file is this:
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
I want to rename this file to:
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using ${parameter%word} (Remove matching suffix pattern):
$ echo "$fn"
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
$ echo "${fn%:*}"
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using cut
$ echo $fn
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
$ echo $fn |cut -d: -f1
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using awk
echo $fn |awk -F : '{print $1}'
more ways...
According to the link here:
This should work:
awk '{old=$0;gsub(/...$/,"",$0);system("mv \""old"\" "$0)}'
provided the file name is given as input.
For eg:
ls -1 NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00|nawk '{old=$0;gsub(/...$/,"",$0);system("mv \""old"\" "$0)}'
Rename file using bash string manipulations:
# Filename needs to be in a variable
file=NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
# Rename file
mv "$file" "${file%???}"
This removes the last three characters from filename.
Using just bash:
fn='NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00'
mv "$fn" "${fn::-3}"
if you have Ruby
echo NBDG6_CD* | ruby -e 'f=gets.chomp;File.rename(f, f[0..-4])'

awk extract "Matthew" from mtjoseph:6/MKTCzMS/YU.:10974:10060:Matthew

How to extract "Matthew" from "mtjoseph:6/MKTCzMS/YU.:10974:10060:Matthew"
echo "mtjoseph:6/MKTCzMS/YU.:10974:10060:Matthew"| awk -F ':' '{print $5}'
I find echo "mtjoseph:6/MKTCzMS/YU.:10974:10060:Matthew"| cut -d: -f5 simpler.
Or, just let the shell do it (assuming bash)
entry="mtjoseph:6/MKTCzMS/YU.:10974:10060:Matthew"
name="${entry##*:}"
echo "$name" # ==> Matthew

How to get extension of a file in shell script

I am trying to get file extension for a file in shell script. But without any luck.
The command I am using is
file_ext=${filename##*.}
and
file_ext = $filename |awk -F . '{if (NF>1) {print $NF}}'
But both of the commands failed to put value in variable file_ext. But when i try
echo $filename |awk -F . '{if (NF>1) {print $NF}}'
It gives me the desired result. I am new to shell script. Please describe the situation what is happening. And also how should I do it?
Thanks.
to get file extension, just use the shell
$ filename="myfile.ext"
$ echo ${filename##*.}
ext
$ file_ext=${filename##*.} #put to variable
$ echo ${file_ext}
ext
Spaces hurt.
Anyway you should do:
file_ext=$(echo $filename | awk -F . '{if (NF>1) {print $NF}}')
[Edit] Better suggestion by Martin:
file_ext=$(printf '%s' "$filename" | awk -F . '{if (NF>1) {print $NF}}')
That will store in $file_ext the output of the command.
You have to be careful when declaring variables.
variable1="string" # assign a string value
variable3=`command` # assign output from command
variable2=$(command) # assign output from command
Notice that you cannot put a space after the variable, because then it gets interpreted as a normal command.

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