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Help with Bash script - set, awk

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# 1  
Old 06-03-2009
Help with Bash script - set, awk
Trying to search a log file for a string, starting from a certain point in the log file. I want to return the number of lines that contain the search string and the total number of lines in the log file. Here's the part of the script I'm having problems with:

Code:
set -- $(awk -v lines=$lineslastrun -v pattern="$searchpattern" '
  BEGIN { count=0 }
  NR > lines
  /pattern/ { count++ }
  END { print NR; print count }
' $logfile)
echo $1
echo $2

The variables are set earlier in the script and are:
$lineslastrun=8
$searchpattern="test data"

The log file looks like this:

Code:
[root ~]$ more testlog
one
two
three
test data
five
six
seven
eight
nine
ten
eleven
twelve
thirteen
fourteen
fifteen
test data
seventeen
eighteen
test data
testdata

When I run the script with 'set -x' I get the following:

Code:
++ awk -v lines=8 -v 'pattern=test data' '
  BEGIN { count=0 }
  NR > lines
  /pattern/ { count++ }
  END { print NR; print count }
' ./testlog
+ set -- nine ten eleven twelve thirteen fourteen fifteen test data seventeen eighteen test data testdata 20 0
+ echo nine
nine
+ echo ten
ten

  • I would like awk to only return the data specified in the END statement but it is returning every line after 8. If it does that then 'echo $1' and 'echo $2' will return Total lines and Number of hits.
  • I can't seem to get awk to search for strings with spaces. The search pattern above should have returned '2' but it's returning '0'. If I search for a pattern without a space it works fine.
Thanks,

Mike G.
# 2  
Old 06-03-2009
The awk code should be:

Code:
awk -v lines=$lineslastrun -v pattern="$searchpattern" '
  BEGIN { count=0 }
  NR > lines && $0 ~ pattern{ count++ }
  END { print NR; print count }
' $logfile

# 3  
Old 06-03-2009
Thank you sir. That worked!
# 4  
Old 07-13-2009
Hello sir,



I have a similar example in which I need to search for a pattern from a particular point of the file. Say for ex,



The file has



asdsf

aaaa

adsajd

asda



Processing



sdfdkjf

aaaa

dflkjd

dfjkhdk

aaaa



I need to search for "aaaa" after the point "Processing" occurs in the file.



Can you please help!



Thanks
# 5  
Old 07-13-2009
Code:
awk '/Processing/{f=1} f && /aaaa/{print;end}' file

# 6  
Old 07-13-2009
Hi, Thanks for the response.
But I get a syntax error

$ echo $pat1
Processing Started
$ echo $pat2
EDI
$ awk '/$pat1/{f=1} f && /$pat2/{print;end}' test.dat
awk: syntax error near line 1
awk: bailing out near line 1
$ awk '/Processing/{f=1} f && /EDI/{print;end}' test.dat
awk: syntax error near line 1
awk: bailing out near line 1
# 7  
Old 07-13-2009
Quote:
Originally Posted by prsshini
Hi, Thanks for the response.
But I get a syntax error

$ echo $pat1
Processing Started
$ echo $pat2
EDI
$ awk '/$pat1/{f=1} f && /$pat2/{print;end}' test.dat
awk: syntax error near line 1
awk: bailing out near line 1
$ awk '/Processing/{f=1} f && /EDI/{print;end}' test.dat
awk: syntax error near line 1
awk: bailing out near line 1
If you're using shell variables, the code should looks like:

Code:
awk '/p1/{f=1} f && /p2/{print;end}' p1=$pat1 p2=$pat2 test.dat

Use nawk or /usr/xpg4/bin/awk on Solaris if you get errors.
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