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Old Unix and Linux 02-02-2011   -   Original Discussion by harjinder
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shell scripting and programmming

hi
i have to grep a certain pattern
like i have

Code:
revision 1.15
date 2011-01-20 author : dpriyank
---------------------------------------
revision 1.10
date 2011-01-10 author : sandeepk
----------------------------------------
revision 1.2
date 2011-01-09 author : tanvi
----------------------------------------
revision 1.1
date 2011-01-02 auhtor : jatin

i just need revision 1.1 and the author name.....what should i do

i tried

Code:
grep -o -e "revision 1.1" -e "author" | awk '{print $3}'

it is giving the answer

Code:
revision 1.1 dpriyank
revision 1.1 sandeepk
revision 1.1 jatin

i only want the revison 1.1 and the name of the concernd author
plzzz help

Last edited by Franklin52; 02-02-2011 at 03:46 AM.. Reason: Please use code tags, thank you
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Old Unix and Linux 02-02-2011   -   Original Discussion by harjinder
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Code:
grep "\b1.1\b" txt -A1 | tr '\n' ' ' | sed 's/date.*://'

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Old Unix and Linux 02-02-2011   -   Original Discussion by harjinder
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Code:
 $ ruby -ne 'print gets.split(":")[-1] if /revision 1.1$/' file

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Old Unix and Linux 02-02-2011   -   Original Discussion by harjinder
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through sed..

Code:
 sed -n '/revision 1.1$/N;s/\n.*:/:/p' inputfile > outfile

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Old Unix and Linux 02-02-2011   -   Original Discussion by harjinder
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revision 1.4
date: 2010-08-15 06:52:24 +0000; author: sandeepk; state: Exp; lines: +1 -1
fixed testcase
----------------------------
revision 1.3
date: 2010-08-15 05:36:28 +0000; author: sandeepk; state: Exp; lines: +2 -0
updated execlevel
----------------------------
revision 1.2
date: 2010-08-15 05:25:02 +0000; author: sandeepk; state: Exp; lines: +15 -3
use name mapping flow
----------------------------
revision 1.1
date: 2010-08-10 10:41:36 +0000; author: sandeepk; state: Exp;
initial check in

revision 1.4
date: 2006-10-31 09:55:13 +0000; author: vkadam; state: Exp; lines: +5 -5
updated
----------------------------
revision 1.3
date: 2006-10-31 08:58:52 +0000; author: vkadam; state: Exp; lines: +1 -1
updated
----------------------------
revision 1.2
date: 2006-10-26 06:19:04 +0000; author: jsaikia; state: Exp; lines: +3 -3
changed exec level to 7.
----------------------------
revision 1.1
date: 2006-10-23 11:24:13 +0000; author: jsaikia; state: Exp;
Testcase on DC block with non-scan design.
----------------------------[/CODE]


i still could not get....this is the actual text from which i want
final answer to be


Code:
revision 1.1      author:jsaikia

revision 1.1 author : sandeepk


this way i have 50 files to edit......i am just taking an example of 2 ...what shud i do for 50 cases like this

Last edited by harjinder; 02-02-2011 at 04:26 AM.. Reason: Please use code tags
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Old Unix and Linux 02-02-2011   -   Original Discussion by harjinder
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Since your recent post's input file is different from the original one you get like that..You could try the below, formatted according to the post#5's input file

Code:
sed -n '/revision 1.1$/N;s/\(.*\)\n.*\(auth.*\); .*/\1 \2/p' inputfile > outfile

The Following User Says Thank You to michaelrozar17 For This Useful Post:
harjinder (02-02-2011)
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Old Unix and Linux 02-02-2011   -   Original Discussion by harjinder
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a more complicated code:

Code:
grep "\b1.1\b" infile -A1 | tr '\n' ' ' | sed -e 's/date.*\(author:.*;\)/\1/' | awk 'BEGIN{FS="[ ;:]"}{print $1,$2,$3":"$5}'

Does this works?
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