bash: awk 打印并打印

Question

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?

Answer 1

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.

Answer 2

您可以尝试以下操作：

awk -F: '/Jenny/ {split($2,a,","); print a[1]}' file 

Answer 3

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 .