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Fetch a value of particular parameter from a log

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# 1  
Old 09-16-2010
Fetch a value of particular parameter from a log
Fetch a value of particular parameter from a log.

for exapmle, i have the below line in the log.

Quote:
[2010:09:16 09:01:55] INFO (Employee Details) Name: prabhu, number:12345, location:india, message:timesout while inseting.
[2010:09:16 09:01:55] INFO (Employee Details) Name: aps, number:67890, location:england, message:timesout while inseting.
now i would like to grep name and fetch only the values of name . here in this case i would like to take out only aps and prabhu.


someone please help me on this.
# 2  
Old 09-16-2010
Code:
# awk -F[,\ ] '{print $7}' LOGFILE
prabhu
aps

# 3  
Old 09-16-2010
Code:
sed 's/^.*Name: \([^,]*\),.*$/\1/' your_file
# or
perl -lne '/.*Name:\s*(\w+),/ and print $1' your_file

tyler_durden
# 4  
Old 09-16-2010
Quote:
Originally Posted by durden_tyler
Code:
sed 's/^.*Name: \([^,]*\),.*$/\1/' your_file

In this case, ^ and $ in sed is useless.

Code:
sed 's/.*Name: \([^,]*\),.*/\1/' infile

# 5  
Old 09-16-2010
Or one could even do:
Code:
sed 's/.*Name: //;s/,.*//' infile

or
Code:
awk -F '[ ,]*' '$0=$7' infile

# 6  
Old 09-16-2010
Code:
ruby -ne 'puts $1 if /Name: (.*?),/' file

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