here is a piece of log
21:36 b05808aa-c6ad-4d30-a334-198ff5726f7c new
22:21 59996d37-9008-4b3b-ab22-340955cb6019 new
21:12 2b41f358-ff6d-418c-a0d3-ac7151c03b78 new
12:36 7ac4995c-ff2c-4717-a2ac-e6870a5670f0 new
i print it by awk '{print $2}' st.log
so i got
b05808aa-c6ad-4d30-a334-198ff5726f7c
59996d37-9008-4b3b-ab22-340955cb6019
2b41f358-ff6d-418c-a0d3-ac7151c03b78
7ac4995c-ff2c-4717-a2ac-e6870a5670f0
now i need to pass it to grep, in this manner
awk '{print $2}' |xargs -i grep -w "pattern from awk" st.log
I need exactly how to pass each founded record from awk to grep. I do not need other solutions, because my task is more complicated, than this piece. Thank you.
With bash and grep:
grep -f <(awk '{print $2}' piece_of_log) st.log
No need for awk:
grep -Ff <(cut -d' ' -f2 log)
It seems you're looking for the replace string option:
-I replace-str
Replace occurrences of replace-str in the initial-arguments with
names read from standard input. Also, unquoted blanks do not
terminate input items; instead the separator is the newline
character. Implies -x and -L 1.
Like this:
awk '{print $2}' | xargs -I{} grep -w {} st.log
Related
I've been trying to print all the instances of a matching pattern from file.
Input file:
{"id":"prod123","a":1.3,"c":"xyz","q":2},
{"id":"prod456","a":1.3,"c":"xyz","q":1}]}
{"id":"prod789","a":1.3,"currency":"xyz","q":2},
{"id":"prod101112","a":1.3,"c":"xyz","q":1}]}
I'd want to print everything between "id":" and ",.
Expected output:
prod123
prod456
prod789
prod101112
I'm using the command
grep -Eo 'id\"\:\"[^"]+"\"\,*' | grep -Eo '^[^"]+'
Am I missing anything here?
What went wrong is the place of the comma in the first grep:
grep -Eo 'id.\:.[^"]+"\,"' inputfile
You need to do something extra for getting the desired substring.
grep -Eo 'id.\:.[^"]+"\,"' inputfile | cut -d: -f2 | grep -Eo '[^",]+'
I used cut, that would be easy for your example input.
cut -d'"' -f4 < inputfile
You have alternatives, like using jq, or
sed -r 's/\{"id":"([^"]*).*/\1/' inputfile
or using awk (solution now like cut but can be changed easy)
awk -F'"' '{print $4}' inputfile
I have this file (serials.txt) containing serial numbers:
S/N:175-1915011190
S/N:244-1920023447
S/N:335-1920101144
S/N:244-1920101149
Using grep or similar tool I want to select all serials NOT starting with '244'
I'm able to select all the '244' with grep -Eo '244-[0-9]*' serials.txt but I want the opposite.
Something like grep -Eo '(^244)-[0-9]*' serials.txt
The output should be (without S/N:)
175-1915011190
335-1920101144
Following awk may help you in same.
awk '!/S\/N:244/' Input_file
EDIT: Above code will give complete line as output if you need starting from serial number to till end in output then following may help you.
awk -F':' '!/S\/N:244/{print $2}' Input_file
EDIT2: Adding a sed solution too here for same.
sed -n '/:244/d;s/.*://;p' Input_file
The -v option on grep would be helpful here, and then cut to remove the leading cruft:
grep -v ':244-' serials.txt | cut -c5-
Here you go, without S/N:
grep -v ':244' serials.txt | cut -d':' -f2
Antigrep for :244, cuts with delimiter : shows field 2.
awk -F':' '$2!~/^244/{print $2}' file
Output of awk '{print $4}' is
b05808aa-c6ad-4d30-a334-198ff5726f7c
59996d37-9008-4b3b-ab22-340955cb6019
2b41f358-ff6d-418c-a0d3-ac7151c03b78
7ac4995c-ff2c-4717-a2ac-e6870a5670f0
I need to grep file st.log by these records. Something like awk '{print $4}' |xargs -i grep -w "pattern from awk" st.log I dont know how to pass pattern correctly?
What about
awk '{print $4}' | grep -F -f - st.log
Credits to Eric Renouf, who noticed that -f - can be used for standard input instead -f <(cat), Note: -f /dev/stdin also works and avoids launching a new process.
or closer to the question to have the output ordered
awk '{print $4}' | xargs -i grep -F {} st.log
maybe -w was not the option OP needed but -F
grep --help
-F, --fixed-strings PATTERN is a set of newline-separated strings
-w, --word-regexp force PATTERN to match only whole words
-w will match only line that contain exactly pattern
examples
grep -w . <<<a # matches
grep -w . <<<ab # doesn't match
grep -F . <<<a # doesn't match
grep -F . <<<a.b # matches
May be something along these lines be helpful
How to process each line received as a result of grep command
awk '{print $4} | while read -r line; do
grep $line st.log
done
I'm trying to write a bash shell script, that opens a certain file CATALOG.dat, containing the following lines, made of both characters and numbers:
event_0133_pk.gz
event_0291_pk.gz
event_0298_pk.gz
event_0356_pk.gz
event_0501_pk.gz
What I wanna do is print the numbers (only the numbers) inside a new file NUMBERS.dat, using something like > ./NUMBERS.dat, to get:
0133
0291
0298
0356
0501
My problem is: how do I extract the numbers from the text lines? Is there something to make the script read just the number as a variable, like event_0%d_pk.gz in C/C++?
A grep solution:
grep -oP '[0-9]+' CATALOG.dat >NUMBERS.dat
A sed solution:
sed 's/[^0-9]//g' CATALOG.dat >NUMBERS.dat
And an awk solution:
awk -F"[^0-9]+" '{print $2}' CATALOG.dat >NUMBERS.dat
There are many ways that you can achieve your result. One way would be to use awk:
awk -F_ '{print $2}' CATALOG.dat > NUMBERS.dat
This sets the field separator to an underscore, then prints the second field which contains the numbers.
Awk
awk 'gsub(/[^[:digit:]]/,"")' infile
Bash
while read line; do echo ${line//[!0-9]}; done < infile
tr
tr -cd '[[:digit:]\n]' <infile
You can use grep command to extract the number part.
grep -oP '(?<=_)\d+(?=_)' CATALOG.dat
gives output as
0133
0291
0298
0356
0501
Or
much simply
grep -oP '\d+' CATALOG.dat
You don't need perl mode in grep for this. BREs can do this.
grep -o '[[:digit:]]\+' CATALOG.dat > NUMBERS.dat
How to cut and overwrite a certain section from a file BASH
Content in dl.txt: "127. www.example.com"
I have tried:
#cat dl.txt|egrep -v "^[0-9]+.[ ]" > dl.txt
#cat dl.txt|egrep "www.example.com" > dl.txt
Could this maybe be done in awk ?
If you mean you want to alter the contents of the file dl.txt and delete "127. ", you can use sed:
sed -i.bak 's/127. //' dl.txt
Then you will see that dl.txt is changed and dl.txt.bak is a backup copy.
Likewise if you want to remove the "www.example.com"
sed -i.bak 's/www.example.com//' dl.txt
Or if you want to delete everything up to, and including the space on each line:
sed -i.bak 's/.* //' dl.txt
Or using awk:
awk '{print $1}' dl.txt
127.
awk '{print $2}' dl.txt
www.example.com
Or you can do it nearly in-place with awk like this, only overwriting the original if the awk is successful:
awk '{print $2}' dl.txt > $$.tmp && mv $$.tmp dl.txt
Below should get rid of 127. before the url, or any ip address for that matter:
sed -i 's/[0-9]\+\.\?\s*//g' dl.txt