This question already has answers here:
Parsing variables from config file in Bash
(7 answers)
Closed 4 years ago.
I have the following bash command:
cat setup.py | grep name=`
This returns the line
name='SOME_PROJECTNAME',
How would I pipe this output from grep to just retrieve SOME_PROJECTNAME?
I have tried
cat setup.py | grep name= | tr -d 'name=','
but this removes characters in SOME_PROJECTNAME.
Use grep lookahead.
$ grep -oP "(?<=name=').*(?=')" setup.py
#bad cat setup.py | grep name= | cut -d= -f2-
cat setup.py | grep name= | cut -d' -f2
Sed may be helpful here.
sed -ne "s/name='\(.*\)'/\1/p" setup.py
The option -n makes sed not print lines by default. Then we replace the entire property line (name='SOME_PROJECTNAME') with the value only (SOME_PROJECTNAME). The p flag in the s/// command makes sed print the line only if the replacement its executed. So, the only line to be printed are the ones where the replacement was done, with the replaced value.
Related
I am just simply trying to grab the commit ID, but not quite sure what I'm missing:
➜ ~ curl https://github.com/microsoft/vscode/releases -s | grep -oE 'microsoft/vscode/commit/(.*?)/hovercard'
microsoft/vscode/commit/ccbaa2d27e38e5afa3e5c21c1c7bef4657064247/hovercard
The only thing I need back from this is ccbaa2d27e38e5afa3e5c21c1c7bef4657064247.
This works just fine on regex101.com and in ruby/python. What am I missing?
If supported, you can use grep -oP
echo "microsoft/vscode/commit/ccbaa2d27e38e5afa3e5c21c1c7bef4657064247/hovercard" | grep -oP "microsoft/vscode/commit/\K.*?(?=/hovercard)"
Output
ccbaa2d27e38e5afa3e5c21c1c7bef4657064247
Another option is to use sed with a capture group
echo "microsoft/vscode/commit/ccbaa2d27e38e5afa3e5c21c1c7bef4657064247/hovercard" | sed -E 's/microsoft\/vscode\/commit\/([^\/]+)\/hovercard/\1/'
Output
ccbaa2d27e38e5afa3e5c21c1c7bef4657064247
The point is that grep does not support extracting capturing group submatches. If you install pcregrep you could do that with
curl https://github.com/microsoft/vscode/releases -s | \
pcregrep -o1 'microsoft/vscode/commit/(.*?)/hovercard' | head -1
The | head -1 part is to fetch the first occurrence only.
I would suggest using awk here:
awk 'match($0,/microsoft\/vscode\/commit\/[^\/]*\/hovercard/){print substr($0,RSTART+24,RLENGTH-34);exit}'
The regex will match a line containing
microsoft\/vscode\/commit\/ - microsoft/vscode/commit/ fixed string
[^\/]* - zero or more chars other than /
\/hovercard - a /hovercard string.
The substr($0,RSTART+24,RLENGTH-34) will print the part of the line starting at the RSTART+24 (24 is the length of microsoft/vscode/commit/) index and the RLENGTH is the length of microsoft/vscode/commit/ + the length of the /hovercard.
The exit command will fetch you the first occurrence. Remove it if you need all occurrences.
You can use sed:
curl -s https://github.com/microsoft/vscode/releases |
sed -En 's=.*microsoft/vscode/commit/([^/]+)/hovercard.*=\1=p' |
head -n 1
head -n 1 is to print the first match (there are 10)grep -o will print (only) everything that matches, including microsoft/ etc.
Your task can not be achieved with Mac's grep. grep -o prints all matching text (compared to default behaviour of printing matching lines), including microsoft/ etc. A grep which implemented perl regex (like GNU grep on Linux) could make use of look ahead/behind (grep -Po '(?<=microsoft/vscode/commit/)[^/]+(?=/hovercard)'). But it's just not available on Mac's grep.
On MacOS you don't have gnu utilities available by default. You can just pipe your output to a simple awk like this:
curl https://github.com/microsoft/vscode/releases -s |
grep -oE 'microsoft/vscode/commit/[^/]+/hovercard' |
awk -F/ '{print $(NF-1)}'
ccbaa2d27e38e5afa3e5c21c1c7bef4657064247
3a6960b964327f0e3882ce18fcebd07ed191b316
f4af3cbf5a99787542e2a30fe1fd37cd644cc31f
b3318bc0524af3d74034b8bb8a64df0ccf35549a
6cba118ac49a1b88332f312a8f67186f7f3c1643
c13f1abb110fc756f9b3a6f16670df9cd9d4cf63
ee8c7def80afc00dd6e593ef12f37756d8f504ea
7f6ab5485bbc008386c4386d08766667e155244e
83bd43bc519d15e50c4272c6cf5c1479df196a4d
e7d7e9a9348e6a8cc8c03f877d39cb72e5dfb1ff
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($0,RSTART+1,RLENGTH-1)}' Input_file
Your grep -Eo ",rds:" is failing for different reasons:
You did not add a " in the string to match
Between the comma and rds you need to match the character.
You are trying to match the comma that can be on the previous line
Your sample input is 2 lines (with a newline in between), perhaps the real input is without the newline.
You want to match until the next double quote.
You can support both input-styles (with/without newline) with
grep -Eo '(,|^)"rds:[^"]*' rdsfile |cut -d'"' -f2
You can do this in one command with
sed -rn 's/.*(,|^)"(rds:[^"]*).*/\2/p' rdsfile
EDIT: Manipulting stdout and not the file is with similar commands:
yourcommand | grep -Eo '(,|^)"rds:[^"]*' |cut -d'"' -f2
# or
yourcommand | sed -rn 's/.*(,|^)"(rds:[^"]*).*/\2/p'
You can also test the original commands with yourcommand > rdsfile.
You might notice that rdsfile is missing data that you have seen on the screen, in that case add 2>&1
yourcommand 2>&1 | grep -Eo '(,|^)"rds:[^"]*' |cut -d'"' -f2
# or
yourcommand 2>&1 | sed -rn 's/.*(,|^)"(rds:[^"]*).*/\2/p'
I'm using bash to look into file and parse the results. Can someone tell me how to use cut/awk to split the string and get desired output by using single command? I can get through individual cut and get the below output (with 2 commands and concatenation) but i want to do using single command instead of two commands.
test.log:
1/98 | (PASSED) com.yahoo.qa.java.projects.stackoverview.questions.Password_01() | 21:20:20
Tried code:
str1=`cat test.log | tail -1 | cut -d '|' -f 1`
str2=`cat test.log | tail -1 | cut -d '|' -f 2 | sed -e 's/com.yahoo.qa.java.projects./''/g'`
str3="${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(/\(\).*/,"",$2)' file
1/98 | (PASSED) stackoverview.questions.Password_01
This question already has an answer here:
Sed command in makefile
(1 answer)
Closed 6 years ago.
I have tried putting the following command in makefile.
#get Local Ip Address
LOCALIP=$(shell ifconfig | grep -Eo 'inet (addr:)?([0-9]*\.){3}[0-9]*' | grep -Eo '([0-9]*\.){3}[0-9]*' | grep -v '127.0.0.1' | awk '{print $1}') &
#get Web Url from User
#read -p "Enter Web Url:" weburl; \
sed -e "\|$LOCALIP $weburl|h; \${x;s|$LOCALIP $weburl||;{g;t};a\\" -e "$LOCALIP $weburl" -e "}" hosts.txt
When I try to execute the command, I expected to get the sed command like following:
sed -e "\|192.168.5.1 www.weburl.com|h; \${x;s|192.168.5.1 www.weburl.com||;{g;t};a\\" -e "192.168.5.1 www.weburl.com" -e "}" hosts.txt
But, I get the following,
sed -e "\|/s/$/OCALIP eburl|h; \" hosts.txt
In Makefiles, variables longer than a single character (i.e. all variables that you're likely to define) needs to be expanded with ${varname}, not $varname. The latter would result in the value of $v concatenated with the string arname, as you discovered.
I won't start to parse the rest of that Makefile as the piping looks a bit questionable.
This question already has answers here:
How to pass command output as multiple arguments to another command
(5 answers)
Closed 3 years ago.
I have this two step bash command:
L=`wc -l testfile | cut -d' ' -f1`
myprogram testfile $L testfile.out
Long story short, myprogram needs the line count as an input.
I want to combine this into one line.
This does not work because using redirect | to - passes stdout stream as a file, not a string.
wc -l testfile | cut -d' ' -f1 | myprogram testfile - testfile.out
Is there a way to combine this into one line?
Use process substitution:
myprogram testfile $(wc -l < testfile) testfile.out
^^^^^^^^^^^^^^^^^^^
This way, wc -l < testfile is evaluated together with the call of the program and you have both commands combined.
Note wc -l < file returns you just the number, so you don't have to do cut or any other thing to clean the output.