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awk get matched line's previous line

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# 1  
Old 08-16-2010
awk get matched line's previous line
hi everyone,

Code:
a
b in
c
d
e
f in
g

output is:
Code:
a
e

so awk search for "in", then print out the matched line's previuos line.
Please advice.
Thanks
# 2  
Old 08-16-2010
try this,

Code:
awk '/in/ { print a } { a = $0}' inputfile

These 2 Users Gave Thanks to pravin27 For This Post:
cola jimmy_y
# 3  
Old 08-16-2010
Quote:
Originally Posted by pravin27
try this,

Code:
awk '/in/ { print a } { a = $0}' inputfile

Thanks
# 4  
Old 08-16-2010
Quote:
Originally Posted by pravin27
try this,

Code:
awk '/in/ { print a } { a = $0}' inputfile

Nice with awk.
Can anyone do the same thing with sed?
# 5  
Old 08-16-2010
Quote:
Originally Posted by cola
Nice with awk.
Can anyone do the same thing with sed?
Something like this?
Code:
sed -n 'N;/in/s/\n.*//p'

# 6  
Old 08-16-2010
Quote:
Originally Posted by Franklin52
Something like this?
Code:
sed -n 'N;/in/s/\n.*//p'

No,
I got the output:
Code:
b in
e

# 7  
Old 08-16-2010
Quote:
Originally Posted by cola
No,
I got the output:
Code:
b in
e

Works fine on my HP-UX system, try <Ctrl-v><Enter> instead of \n:
Code:
sed -n 'N;/in/s/^M.*//p'

This User Gave Thanks to Franklin52 For This Post:
jimmy_y
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