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
Related
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.
I am trying to read/grep a particular word or content that is before a period (.).
e.g. file1 has abinaya.ashok and I want to grep whatever is before the period (.) without hardcoding anything.
if I try
grep \.\ file1
it gives abinaya.ashok.
I've tried: grep\*\.\ file1
it doesn't give anything.Can we find it using grep commands or should we do it only using awk command? Any thoughts?
Using GNU grep for PCRE regex (for non-greedy and positive look-ahead), you can do:
echo 'abinaya.ashok' | grep -oP '.*?(?=\.)'
abinaya
Using awk:
echo 'abinaya.ashok' | awk -F\. '{print $1}'
abinaya
Check the following simple examples.
Including the dot:
$ echo abinaya.ashok | grep -o '.*[.]'
abinaya.
Without the dot:
$ echo abinaya.ashok | grep -o '^[^.]\+'
abinaya
Hope I understand you correctly:
sed -n 's/\..*//p' file1 | grep whatever
sed expression will print only part before dot (lines without dot are not printed).
Now use grep to search what you need.
I am trying print the first field of the first row of an output. Here is the case. I just need to print only SUSE from this output.
# cat /etc/*release
SUSE Linux Enterprise Server 11 (x86_64)
VERSION = 11
PATCHLEVEL = 2
Tried with cat /etc/*release | awk {'print $1}' but that print the first string of every row
SUSE
VERSION
PATCHLEVEL
Specify NR if you want to capture output from selected rows:
awk 'NR==1{print $1}' /etc/*release
An alternative (ugly) way of achieving the same would be:
awk '{print $1; exit}'
An efficient way of getting the first string from a specific line, say line 42, in the output would be:
awk 'NR==42{print $1; exit}'
Specify the Line Number using NR built-in variable.
awk 'NR==1{print $1}' /etc/*release
try this:
head -1 /etc/*release | awk '{print $1}'
df -h | head -4 | tail -1 | awk '{ print $2 }'
Change the numbers to tweak it to your liking.
Or use a while loop but thats probably a bad way to do it.
You could use the head instead of cat:
head -n1 /etc/*release | awk '{print $1}'
sed -n 1p /etc/*release |cut -d " " -f1
if tab delimited:
sed -n 1p /etc/*release |cut -f1
Try
sed 'NUMq;d' /etc/*release | awk {'print $1}'
where NUM is line number
ex. sed '1q;d' /etc/*release | awk {'print $1}'
awk, sed, pipe, that's heavy
set `cat /etc/*release`; echo $1
the most code-golfy way i could think of to print first line only in awk :
awk '_{exit}--_' # skip the quotations and make it just
# awk _{exit}--_
#
# if u're feeling adventurous
first pass through exit block, "_" is undefined,
so it fails and skips over for row 1.
then the decrementing of the same counter will make
it "TRUE" in awk's eyes (anything not empty string
or numeric zero is considered "true" in their agile boolean sense). that same counter also triggers default action of print for row 1.
—- incrementing… decrementing… it's same thing,
merely direction and sign inverted.
then finally, at start of row 2, it hits criteria to
enter the action block, which instructs it to instantly
exit, thus performing essentially the same functionality as
awk '{ print; exit }'
… in a slightly less verbose manner. For a single line print, it's not even worth it to set FS to skip the field splitting part.
using that concept to print just 1st row 1st field :
awk '_{exit} NF=++_'
awk '_++{exit} NF=_'
awk 'NR==1&&NF=1' file
grep -om1 '^[^ ]\+' file
# multiple files
awk 'FNR==1&&NF=1' file1 file2
You can kill the process which is running the container.
With this command you can list the processes related with the docker container:
ps -aux | grep $(docker ps -a | grep container-name | awk '{print $1}')
Now you have the process ids to kill with kill or kill -9.
I need to get a list of matches with grep including filename and line number but without the match string
I know that grep -Hl will give only file names and grep -Hno will give filename with only matching string. But those not ideal for me. I need to get a list without match but with line no. For this grep -Hln doesn't work. I tried with grep -Hn 'pattern' | cut -d " " -f 1 But it doesn't cut the filename and line no properly.
awk can do that in single command:
awk '/pattern/ {print FILENAME ":" NR}' *.txt
You were pointing it well with cut, only that you need the : field separator. Also, I think you need the first and second group. Hence, use:
grep -Hn 'pattern' files* | cut -d: -f1,2
Sample
$ grep -Hn a a*
a:3:are
a:10:bar
a:11:that
a23:1:hiya
$ grep -Hn a a* | cut -d: -f1,2
a:3
a:10
a:11
a23:1
I guess you want this, just line numbers:
grep -nh PATTERN /path/to/file | cut -d: -f1
example output:
12
23
234
...
Unfortunately you'll need to use cut here. There is no way to do it with pure grep.
Try
grep -RHn Studio 'pattern' | awk -F: '{print $1 , ":", $2}'
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