I need to grep some pattern and further i need to print some output within that. Currently I am using the below command which is working fine. But I like to eliminate using multiple pipe and want to use single awk command to achieve the same output. Is there a way to do it using awk?
root#Server1 # cat file
Jenny:Mon,Tue,Wed:Morning
David:Thu,Fri,Sat:Evening
root#Server1 # awk '/Jenny/ {print $0}' file | awk -F ":" '{ print $2 }' | awk -F "," '{ print $1 }'
Mon
I want to get this output using single awk command. Any help?
You can try something like:
awk -F: '/Jenny/ {split($2,a,","); print a[1]}' file
Try this
awk -F'[:,]+' '/Jenny/{print $2}' file.txt
It is using muliple -F value inside the [ ]
The + means one or more since it is treated as a regex.
For this particular job, I find grep to be slightly more robust.
Unless your company has a policy not to hire people named Eve.
(Try it out if you don't understand.)
grep -oP '^[^:]*Jenny[^:]*:\K[^,:]+' file
Or to do a whole-word match:
grep -oP '^[^:]*\bJenny\b[^:]*:\K[^,:]+' file
Or when you are confident that "Jenny" is the full name:
grep -oP '^Jenny:\K[^,:]+' file
Output:
Mon
Explanation:
The stuff up until \K speaks for itself: it selects the line(s) with the desired name.
[^,:]+ captures the day of week (in this case Mon).
\K cuts off everything preceding Mon.
-o cuts off anything following Mon.
Related
I am trying to capture the string between 2 specific letters/words using sed/awk. This is what I am trying to do:
The input is a file test.log containing
Owner: CN=abc.samplecerrt.com,o=IN,DC=com
Owner: CN=abc1.samplecerrt.com,o=IN,DC=com
I want to extract only "CN=abc.samplecerrt.com"
I tried
sed 's/.*CN=\(.*\),.*/\1/p' test.log >> result.log
But this returns "abc.samplecerrt.com,o=IN,DC=com"
How do I go about this?
test file:
$ cat logs.txt
CN=abc.samplecerrt.com,o=IN,DC=com Owner: CN=abc1.samplecerrt.com,o=IN,DC=com
command and output:
$ grep -oP 'CN=(?:(?!CN=).)*?.com' logs.txt
CN=abc.samplecerrt.com
CN=abc1.samplecerrt.com
This might work for you (GNU sed):
sed -n 's/.*\(CN=[^,]*\).*/\1/p' file
Or:
sed 's/.*\(CN=[^,]*\).*/\1/p;d' file
The first turns off implicit printing -n so as to act like grep.
Matches and captures the string CN= followed by zero or more non-comma characters and prints the captured group \1 if a match is made.
The second solution is much the same except it deletes all lines and only prints the captured group as above.
With awk you can get the field where is the string you need. For it, you can set FS=:|, Now if you run
awk -v FS=":|," '{print $2}' file
CN=abc.samplecerrt.com
CN=abc1.samplecerrt.com
you get the field. But you only want one, so
awk -v FS=":|," '$2 !~ /abc1/ {print $2}' file
CN=abc.samplecerrt.com
I have CSV columns that I'm working with:
info,example-string,super-example-string,otherinfo
I would like to get:
example-string super example string
Right now, I'm running the following command:
awk -F ',' '{print $3}' | sed "s/-//g"
But, then I have to paste the lines together to combine $2 and $3.
Is there anyway to do something like this?
awk -F ',' '{print $2" "$3}' | sed "s/-//g"
Except, where the sed command is only performed on $3 and $2 stays in place? I'm just concerned later on if the lines don't match up, the data could be misaligned.
Please note: I need to keep the pipe for the SED command. I just used a simple example but I end up running a lot of commands after that as well.
Try:
$ awk -F, '{gsub(/-/," ",$3); print $2,$3}' file
example-string super example string
How it works
-F,
This tells awk to use a comma as the field separator.
gsub(/-/," ",$3)
This replaces all - in field 3 with spaces.
print $2,$3
This prints fields 2 and 3.
Examples using pipelines
$ echo 'info,example-string,super-example-string,otherinfo' | awk -F, '{gsub(/-/," ",$3); print $2,$3}'
example-string super example string
In a pipeline with sed:
$ echo 'info,example-string,super-example-string,otherinfo' | awk -F, '{gsub(/-/," ",$3); print $2,$3}' | sed 's/string/String/g'
example-String super example String
Though best solution will be either use a single sed or use single awk. Since you have requested to use awk and sed solution so providing this. Also considering your actual data will be same as shown sample Input_file.
awk -F, '{print $2,$3}' Input_file | sed 's/\([^ ]*\)\([^-]*\)-\([^-]*\)-\([^-]*\)/\1 \2 \3 \4/'
Output will be as follows.
example-string super example string
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
I'd like to know how cat passwd | awk -F':' '{printf $1}' works. cat /etc/passwd is a list of users with ID and folders from root to the current user (I don't know if it has something to do with cat passwd). -F is some kind of input file and {printf $1} is printing the first column. That's what I've search so far but seems confusing to me.
Can anyone help me or explain to me if it's right or wrong, please?
This is equivalent to awk -F: '{print $1}' passwd. The cat command is superfluous as all it does is read a file.
The -F option determines the field separator for awk. The quotes around the colon are also superfluous since colon is not special to the shell in this context. The print invocation tells awk to print the first field using $1. You are not passing a format string, so you probably mean print instead of printf.
Suppose I have a file input.txt with few columns and few rows, the first column is the key, and a directory dir with files which contain some of these keys. I want to find all lines in the files in dir which contain these key words. At first I tried to run the command
cat input.txt | awk '{print $1}' | xargs grep dir
This doesn't work because it thinks the keys are paths on my file system. Next I tried something like
cat input.txt | awk '{system("grep -rn dir $1")}'
But this didn't work either, eventually I have to admit that even this doesn't work
cat input.txt | awk '{system("echo $1")}'
After I tried to use \ to escape the white space and the $ sign, I came here to ask for your advice, any ideas?
Of course I can do something like
for x in `cat input.txt` ; do grep -rn $x dir ; done
This is not good enough, because it takes two commands, but I want only one. This also shows why xargs doesn't work, the parameter is not the last argument
You don't need grep with awk, and you don't need cat to open files:
awk 'NR==FNR{keys[$1]; next} {for (key in keys) if ($0 ~ key) {print FILENAME, $0; next} }' input.txt dir/*
Nor do you need xargs, or shell loops or anything else - just one simple awk command does it all.
If input.txt is not a file, then tweak the above to:
real_input_generating_command |
awk 'NR==FNR{keys[$1]; next} {for (key in keys) if ($0 ~ key) {print FILENAME, $0; next} }' - dir/*
All it's doing is creating an array of keys from the first file (or input stream) and then looking for each key from that array in every file in the dir directory.
Try following
awk '{print $1}' input.txt | xargs -n 1 -I pattern grep -rn pattern dir
First thing you should do is research this.
Next ... you don't need to grep inside awk. That's completely redundant. It's like ... stuffing your turkey with .. a turkey.
Awk can process input and do "grep" like things itself, without the need to launch the grep command. But you don't even need to do this. Adapting your first example:
awk '{print $1}' input.txt | xargs -n 1 -I % grep % dir
This uses xargs' -I option to put xargs' input into a different place on the command line it runs. In FreeBSD or OSX, you would use a -J option instead.
But I prefer your for loop idea, converted into a while loop:
while read key junk; do grep -rn "$key" dir ; done < input.txt
Use process substitution to create a keyword "file" that you can pass to grep via the -f option:
grep -f <(awk '{print $1}' input.txt) dir/*
This will search each file in dir for lines containing keywords printed by the awk command. It's equivalent to
awk '{print $1}' input.txt > tmp.txt
grep -f tmp.txt dir/*
grep requires parameters in order: [what to search] [where to search]. You need to merge keys received from awk and pass them to grep using the \| regexp operator.
For example:
arturcz#szczaw:/tmp/s$ cat words.txt
foo
bar
fubar
foobaz
arturcz#szczaw:/tmp/s$ grep 'foo\|baz' words.txt
foo
foobaz
Finally, you will finish with:
grep `commands|to|prepare|a|keywords|list` directory
In case you still want to use grep inside awk, make sure $1, $2 etc are outside quote.
eg. this works perfectly
cat file_having_query | awk '{system("grep " $1 " file_to_be_greped")}'
// notice the space after grep and before file name