Grep or sed - printing line only with exact match


 
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# 1  
Old 02-18-2015
Grep or sed - printing line only with exact match

Hello.

In my script, some command return :
Code:
q | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
  | kernel-default-base           | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default-base           | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
  | kernel-default-base-debuginfo | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default-base-debuginfo | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
  | kernel-default-debuginfo      | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default-debuginfo      | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
  | kernel-default-debugsource    | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default-debugsource    | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
v | kernel-default-devel          | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
v | kernel-default-devel          | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
kernel-default-devel
v | kernel-default-devel          | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
v | kernel-default-devel          | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard

If my pattern is "kernel-default", I want only this

Code:
q | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard

I want to print the full line.

I have tried :
Code:
grep -o

Code:
grep -w

Code:
grep  "/bxxxx/b"

Code:
grep  "/<xxxx/>"

Any help is welcome.
# 2  
Old 02-18-2015
Hi,
Try:
Code:
grep "[[:blank:]]kernel-default[[:blank:]]"

Regards.
This User Gave Thanks to disedorgue For This Post:
# 3  
Old 02-18-2015
Quote:
Originally Posted by disedorgue
Hi,
Try:
Code:
grep "[[:blank:]]kernel-default[[:blank:]]"

Regards.
Thank you very much
# 4  
Old 02-19-2015
Hello jcdole,

Following may also help you in same.
Code:
awk -F"|" '($2 ~ /^ kernel-default [[:space:]]+/){print $0}'  Input_file

Output will be as follows.
Code:
q | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | x86_64 | openSUSE-13.2-Kernel_stable_standard
  | kernel-default                | package | 3.19.0-1.1.g8a7d5f9 | i586   | openSUSE-13.2-Kernel_stable_standard

Thanks,
R. Singh
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