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Grep and substr

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# 1  
Old 11-09-2009
Grep and substr
Hi,

A quick one, I abstract the last line from a file which hasa string PID in itie.

grep PID ~/bug_tool.log | tail -1

result
PID : 25803 TID : 47983424956736PROC : db2sysc 1


but any ideas on the best way to grab only the string 25803

Thanks
# 2  
Old 11-09-2009
try this ...

Code:
grep PID ~/bug_tool.log | tail -1 | awk -F "[: ]" '{print $4}'

it depends of the blank before the word PID, try to replace $4 by $5
Last edited by protocomm; 11-09-2009 at 09:00 AM..
# 3  
Old 11-09-2009
thanks but returns blank.. i was thinking a substr wrapped around the grep but it doesnt run for me
# 4  
Old 11-09-2009
one other way also.,
Code:
echo 'PID : 25803 TID : 47983424956736PROC : db2sysc 1' | grep -E -o 'PID : [0-9]+' | grep -o -E '[0-9]+'

# 5  
Old 11-09-2009
Code:
echo "PID : 25803 TID : 47983424956736PROC : db2sysc 1" | awk '{print $3}'
25803

Or in one awk statement:
Code:
awk '/PID/{z=$3}END{print z}' ~/bug_tool.log
25803

Last edited by Scrutinizer; 11-09-2009 at 09:07 AM..
# 6  
Old 11-09-2009
Also, you can try the simple 'cut' command as used below -

Code:
grep PID ~/bug_tool.log | tail -1 | cut -d":" -f2

That would return - 25803 TID. To cut out TID you can use

Code:
grep PID ~/bug_tool.log | tail -1 | cut -d":" -f2 | tr -s " " ":" | cut -d":" -f1

tr -s " " ":" - give a space in between the first quotes.
# 7  
Old 11-09-2009
Quote:
Originally Posted by ggs
Code:
grep PID ~/bug_tool.log | tail -1 | cut -d":" -f2 | tr -s " " ":" | cut -d":" -f1

all these can be done with one single awk command like what scrutinizer had posted.
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