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
Related
I am trying to use grep to filter out the RDS snapshot identifier from the rds describe-db-snapshots command output below:
"arn:aws:rds:ap-southeast-1:123456789:snapshot:rds:apple-pie-2018-05-06-17-12",
"rds:apple-pie-2018-05-06-17-12",
how to return the exact output as in
rds:apple-pie-2018-05-06-17-12
tried using
grep -Eo ",rds:"
but not able to
Following awk may also help you on same.
awk 'match($0,/^"rds[^"]*/){print substr($0,RSTART+1,RLENGTH-1)}' Input_file
Your grep -Eo ",rds:" is failing for different reasons:
You did not add a " in the string to match
Between the comma and rds you need to match the character.
You are trying to match the comma that can be on the previous line
Your sample input is 2 lines (with a newline in between), perhaps the real input is without the newline.
You want to match until the next double quote.
You can support both input-styles (with/without newline) with
grep -Eo '(,|^)"rds:[^"]*' rdsfile |cut -d'"' -f2
You can do this in one command with
sed -rn 's/.*(,|^)"(rds:[^"]*).*/\2/p' rdsfile
EDIT: Manipulting stdout and not the file is with similar commands:
yourcommand | grep -Eo '(,|^)"rds:[^"]*' |cut -d'"' -f2
# or
yourcommand | sed -rn 's/.*(,|^)"(rds:[^"]*).*/\2/p'
You can also test the original commands with yourcommand > rdsfile.
You might notice that rdsfile is missing data that you have seen on the screen, in that case add 2>&1
yourcommand 2>&1 | grep -Eo '(,|^)"rds:[^"]*' |cut -d'"' -f2
# or
yourcommand 2>&1 | sed -rn 's/.*(,|^)"(rds:[^"]*).*/\2/p'
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 $2}' file
I'm using bash to look into file and parse the results. Can someone tell me how to use cut/awk to split the string and get desired output by using single command? I can get through individual cut and get the below output (with 2 commands and concatenation) but i want to do using single command instead of two commands.
test.log:
1/98 | (PASSED) com.yahoo.qa.java.projects.stackoverview.questions.Password_01() | 21:20:20
Tried code:
str1=`cat test.log | tail -1 | cut -d '|' -f 1`
str2=`cat test.log | tail -1 | cut -d '|' -f 2 | sed -e 's/com.yahoo.qa.java.projects./''/g'`
str3="${str1} | ${str2}"
Expected:
1/98 | (PASSED) stackoverview.questions.Password_01
Since this is a simple substitution on an individual line it's better suited to sed than awk and not at all appropriate for cut:
$ sed 's/\(.*| [^ ]* \)com\.yahoo\.qa\.java\.projects\.\([^(]*\).*/\1\2/' file
1/98 | (PASSED) stackoverview.questions.Password_01
Following single awk may help you in same.
awk 'END{sub(/com\.yahoo\.qa\.java\.projects\./,"",$4);print $1,$2,$3,$4}' Input_file
OR for all kind of awks following may help you in same too.(As per SIR ED's suggestions):
awk '{value=$0} END{split(value, a," ");sub(/com.yahoo.qa.java.projects\./,"",a[4]);print a[1],a[2],a[3],a[4]}' Input_file
Using awk
$ awk -F "com[.]yahoo[.]qa[.]java[.]projects[.]" 'sub(/\(\).*/,"",$2)' file
1/98 | (PASSED) stackoverview.questions.Password_01
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
Having trouble with grepping and cutting at the same time
I have a file test.txt.
Inside the file is this syntax
File: blah.txt Location: /home/john/Documents/play/blah.txt
File: testing.txt Location /home/john
My command is ./delete -r (filename), say filename is blah.txt.
How would i search test.txt for blah.txt and cut the /home/john/Documents/play/blah.txt out and put it in a variable
grep -P "^File: blah\.txt Location: .+" test.txt | cut -d: -f3
Prefer always to involse as less as possible external command for your task.
You can achive what you want using single awk command:
awk '/^File: blah.txt/ { print $4 }' test.txt
Try this one ;)
filename=$(grep 'blah.txt' test.txt | grep -oP 'Location:.*' | grep -oP '[^ ]+$')
./delete $filename