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Shell Programming and Scripting

Slash delimiter

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# 1  
Old 12-11-2012
Slash delimiter
I have a log file which on field $11 there is sth like: /A/B/C and I want to extract the last part which is C.
for space delimiter I used:
Code:
awk '{print $2 " " $5 ";" $7 ";" $9 ";" $11}'

and had no problem.
but dont know how to extract the 3rd part of the slash delimiter.

That would be great if someone can help me in an easy way as I am just a beginner.
Last edited by Scrutinizer; 12-11-2012 at 06:26 AM.. Reason: Please select the segment before pressing the code button
# 2  
Old 12-11-2012
Try using sub(/.*\//,"",$11); before the print statement..
# 3  
Old 12-11-2012
OR..

Code:
awk '{n=split($11,P,"/");print $2 " " $5 ";" $7 ";" $9 ";" P[n]}'

This User Gave Thanks to pamu For This Post:
frhling
# 4  
Old 12-12-2012
Hello
1- Thanks alot.
2- Could you please explain what n and P are?
# 5  
Old 12-12-2012
Quote:
Originally Posted by frhling
Could you please explain what n and P are?
Here we are spiting that string into multiple fields.
n= Total number of fields.
P= This is an array which stores those fields from index as 1 to n.

Code:
$ echo "first-second-third" | awk '{n=split($0,P,"-");print n;for(i=1;i<=n;i++){print P[i],i}}'
3  # prints n here- it has 3 number of fields.
first 1
second 2
third 3

I hope this helps

pamu
This User Gave Thanks to pamu For This Post:
frhling
# 6  
Old 12-12-2012
try this to take c out of your pattern as required:
Code:
echo '/A/B/C' | sed 's/^\/.*\///'

# 7  
Old 12-12-2012
Really Thanks.
I dont know how to mark as :SOLVED but it is solvedSmilie
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