how to find a word in a file that appears next to a given keyword


 
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# 1  
Old 09-02-2010
how to find a word in a file that appears next to a given keyword

Hi Experts,

I have a file which contains some text. i need to print the word next to a given keyword. Please help.

Ex: test.txt
Code:
 
=====================
NEXT HOST
=====================
 AEADBAS001
access-list 1 permit xxxxxxxxxxxxxx
ip access-list extended BLA_Outgoing_Filter
access-list 1 permit xxxxxxxxxxxxxx
access-list 2 permit xxxxxxxxxxxxxx

If i use the keyword "extended" it should return BLA_Outgoing_Filter

Moderator's Comments:
Mod Comment Having 25 posts you should be familiar using code tags.

Last edited by zaxxon; 09-02-2010 at 07:53 AM..
# 2  
Old 09-02-2010
Code:
[pelccbar@itchy tmp]$ perl -nle 'print $& if /(?<=extended )\w+/' file
BLA_Outgoing_Filter

# 3  
Old 09-02-2010
Code:
sed -n 's/.* extended //p' infile

Code:
awk '$3=="extended"{print $4}' infile

# 4  
Old 09-02-2010
hi Scrutinizer! sed command works. but it shows the rest of the line. i want to retrieve just the next word. it might be separated by "_"s or "."s.

Thanks

---------- Post updated at 07:33 AM ---------- Previous update was at 07:32 AM ----------

Quote:
Originally Posted by bartus11
Code:
[pelccbar@itchy tmp]$ perl -nle 'print $& if /(?<=extended )\w+/' file
BLA_Outgoing_Filter


Hi Bartus11. your command works. but it only shows the prefix if "." was found in the word

---------- Post updated at 09:47 AM ---------- Previous update was at 07:33 AM ----------

i found a workaround to get just the word after the keyword. still i would like to see how this can be done through sed.

Still this only works only if there is one white space between the keyword and the next word. how to eliminate all the white spaces between the keyword and the next word?
# 5  
Old 09-02-2010
Code:
sed 's/.* extended /_/' test.txt

What when more then once is word "extended"? My script do it only once.
with awk
Code:
awk -F " extended " 'OFS="_"; {print $0}' test.txt

# 6  
Old 09-02-2010
Quote:
Originally Posted by mwrg
Hi Experts,

I have a file which contains some text. i need to print the word next to a given keyword. Please help.

Ex: test.txt
Code:
 
=====================
NEXT HOST
=====================
 AEADBAS001
access-list 1 permit xxxxxxxxxxxxxx
ip access-list extended BLA_Outgoing_Filter
access-list 1 permit xxxxxxxxxxxxxx
access-list 2 permit xxxxxxxxxxxxxx

If i use the keyword "extended" it should return BLA_Outgoing_Filter

Moderator's Comments:
Mod Comment Having 25 posts you should be familiar using code tags.

Hello,

Code:
gaurav@localhost:~/unix$ cat test2.txt 
=====================
NEXT HOST
=====================
 AEADBAS001
access-list 1 permit xxxxxxxxxxxxxx
ip access-list extended BLA_Outgoing_Filter
access-list 1 permit xxxxxxxxxxxxxx
access-list 2 permit xxxxxxxxxxxxxx
gaurav@localhost:~/unix$ perl -wln -e 'print $1 if /.*\sextended\s(.*)\s?.*$/' test2.txt 
BLA_Outgoing_Filter
gaurav@localhost:~/unix$ perl -wln -e 'print $1 if /.*\s1\s(.*)\s?.*$/' test2.txt 
permit xxxxxxxxxxxxxx
permit xxxxxxxxxxxxxx
gaurav@localhost:~/unix$

Regards,
Gaurav.
# 7  
Old 09-02-2010
Quote:
Originally Posted by mwrg
hi Scrutinizer! sed command works. but it shows the rest of the line. i want to retrieve just the next word. it might be separated by "_"s or "."s.

Thanks
Hi, that would be:
Code:
awk -F '[ \t._]*' '$3=="extended"{print $4}' infile

or
Code:
 sed -n 's/.*[ \t._]extended[ \t._]*\([^ \t._]*\).*/\1/p' infile

Note that this will return "BLA" since underscore is now a separator...
(if on Solaris use nawk or /usr/bin/xpg4/awk and /usr/bin/xpg4/sed)

Last edited by Scrutinizer; 09-02-2010 at 03:50 PM..
 
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