Printing previous lines !!

[christiniaroelin]

Hi all,
I have the following code:
awk ' { str[NR] = $0 } END {
if ( ( $0 ~ / files found/) && ( $2< '$min_file'))
{
print $2 " files found : WARNING : EXPECTED MINIMUM NUMBER OF FILES " "'"$min_file"'" "\n\n"
print NR
for ( i=NR; i>(NR-5) ; i-- )
print str

}
if ( ( $0 ~ / *usr /) && ( $3 < 9)) {
print $0 "\n WARNING : HEADER NLOD FOUND FOR THE ABOVE LISTING HAVING ONLY : " $3 " BYTES \n\n"
}
}' < outnlod.dat

Im trying to print the previous 5 lines from -files found- record and which meets the criteria of min files found.It all worked fine until i added the for statement to print the previous 5 lines that meets the 'if' criteria.
Please let me know what im missing. I'm a beginner and learning awk by experimenting with examples as above. Your help in my awk pursuit is much appreciated.

Thanks again
Roelin

 

[PHV]

You may try this revision of your code:
awk '
{ str[NR] = $0 }
/ files found/{
if ( $2< '$min_file' )
{
print $2 " files found : WARNING : EXPECTED MINIMUM NUMBER OF FILES " "'"$min_file"'" "\n\n"
print NR
for ( i=NR; i>(NR-5) ; i-- )
print str
}
}
/ *usr /{ if ( $3 < 9))
print $0 "\n WARNING : HEADER NLOD FOUND FOR THE ABOVE LISTING HAVING ONLY : " $3 " BYTES \n\n"
}
' < outnlod.dat

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ222-2244

 

[futurelet]

christiniaroelin, if you want to learn the right way to write Awk programs:

Don't do this:
if ( ( $0 ~ / files found/) && ( $2< '$min_file'))

Instead, make a file containing only the Awk program and run it this way:
[tt]
awk -v min_file="$min_file" -f prog.awk infile >outfile
[/tt]
The line in your program should be changed to
if ( ( $0 ~ / files found/) && ( $2 < min_file))

 

[christiniaroelin]

Thanks a lot PHV and futurlet..!!
Appreciate your time and help!!

Regrads,
Roelin!!