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Shell Programming and Scripting

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# 1  
Old 04-30-2009
output required
Hi

I have the following output

Message man amm (9196) is calling
Hello & Alert man amtrr (9197) is stopped
Find amfi (19198) is cancelled
Engine Item aea (19203) is notified
Engine Item2 aea2 (19204) is notified

I need to print only the column which contains amm,amtrr,amfi,aea & aea2

in awk $3 does not give 3rd line i.e amfi.... The clue is the items i need to print starts with 'a'
# 2  
Old 04-30-2009
Assuming the pattern is always before the field between the parentheses:

Code:
sed 's/.* \(.*\) (.*/\1/' file

Last edited by Franklin52; 04-30-2009 at 06:18 PM.. Reason: literal correction
# 3  
Old 04-30-2009
no,

I do have brackets before the pattern
# 4  
Old 04-30-2009
Quote:
Originally Posted by PrasannaKS
no,

I do have brackets before the pattern
That's what I mean, have you tried the command?
Last edited by Franklin52; 04-30-2009 at 06:36 PM..
# 5  
Old 04-30-2009
i also have down the file like

Req Har omrh (2254) is terminating
MR Rect (MRRT) omrrt (2255) is concluding
DR Rect (DRRT) omrdt (2256) is concluding
# 6  
Old 04-30-2009
Have you tried the command?
# 7  
Old 04-30-2009
yes, the first half of the output is empty and then one line with the correct word printed and the blank output.

In simple terms, I need to print the word which starts with a letter 'o' for each line.

<---Only one word in each line which starts with letter 'o' . That word could be anywhere in the line i.e $1 or $2 or $3 or $4--->
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