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Bash - Extracting whole word containing substring


 
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# 1  
Old 04-02-2013
Bash - Extracting whole word containing substring

hello.
I get this text when using some command :
Code:
S | Name            | Type    | Version | Arch | Repository  
--+-----------------+---------+---------+------+-------------
  | AdobeReader_enu | package | 9.5.4-1 | i486 | zypper_local

I need to get "AdobeReader_enu" from the the pattern "Ado"
something like :

FILE_NAME=$( some_command | grep -o "Ado" )

Any help is welcome.
# 2  
Old 04-02-2013
Code:
FILE_NAME=$( some_command | grep -o "Ado[^[:space:]]*" )

This User Gave Thanks to Yoda For This Post:
# 3  
Old 04-02-2013
grep -o "Ado[^[:space:]]*" works for the example but not for a string like xyzAdobeReader_enu.
# 4  
Old 04-02-2013
Quote:
Originally Posted by Yoda
Code:
FILE_NAME=$( some_command | grep -o "Ado[^[:space:]]*" )

Great.
Please can you explain --> Ado[^[:space:]]*

Thank you for helping.
# 5  
Old 04-02-2013
[:space:] is a character class that represents tab, newline, vertical tab, form feed, carriage return, and space.

^ represents character not in the list.

* represents zero or more occurrences.
# 6  
Old 04-02-2013
Quote:
Originally Posted by Yoda
Code:
FILE_NAME=$( some_command | grep -o "Ado[^[:space:]]*" )

Quote:
Originally Posted by rdrtx1
grep -o "Ado[^[:space:]]*" works for the example but not for a string like xyzAdobeReader_enu.
exact.

Code:
S | Name            | Type    | Version | Arch | Repository  
--+-----------------+---------+---------+------+-------------
  | AdobeReader_enu | package | 9.5.4-1 | i486 | zypper_local

some_command | grep -o "Rea[^[:space:]]*"

Return :

Code:
Reader_enu

So how to improve ?

---------- Post updated at 23:20 ---------- Previous update was at 23:18 ----------

Quote:
Originally Posted by Yoda
[:space:] is a character class that represents tab, newline, vertical tab, form feed, carriage return, and space.

^ represents character not in the list.

* represents zero or more occurrences.
Thank you
# 7  
Old 04-02-2013
try:
Code:
grep -o "[^[:space:]]*Rea[^[:space:]]*"

or
Code:
sed -n '/Rea/s/.*\b\(.*Rea\w*\).*/\1/p'

which will also return the word between boundary and end of word. (in case word is wrapped by white space or non white space chars)

Last edited by rdrtx1; 04-02-2013 at 06:29 PM..
This User Gave Thanks to rdrtx1 For This Post:

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