Stripping certain lines from a file

[Dave Walley]

Hi.

I have a requirement to select from a file all lines following the first line with the numbers 0 and 8 in positions 61 and 62 respectively.


I have tried some sed functions
sed -n '.{60}08' filename
without success.


Any suggestions

Dave

[Steve Sauve]

You could use "cut -c 61,62" and an it/then routine to check the results of the cut. Then write the result to a new file if it doesn't equal the 08. That would give you a new file with only the lines you want.

Hope this helps,
Steve

[James R. Ferguson]

Dave:

cat /tmp/in|cut -c61-62|grep "08" > /tmp/out

...JRF...

[Emilio Sierra]

Test with
grep '^.{60}08' filename

[Jaroslaw Tomkiewicz]

Hi Dave,

try this command:
awk '{if (substr($0,60,2)=="08") print }' name_of_your_file


regards

Jaroslaw

[Tom Danzig]

Try:

awk '{if(NR>1){if(substr($0,61,2)=="08"){print $0}}}' input_file_name

[Dave Walley]

Thank you all for your suggestions but I require every line from the first occurence of 08 in 61 and 62 regardless of what is in columns 61 and 62 thereafter.

[Tom Danzig]

awk '{if(substr($0,61,2)=="08"){flag=1}{if(flag==1){print $0}}' input_file_name

[Tom Danzig]

Sorry. My last post had a syntax error:

awk '{if(substr($0,61,2)=="08"){flag=1}{if(flag==1){print $0}}}' input_file_name

[Emilio Sierra]

a=`grep -n '^.{9}08' filename | cut -d':' -f1|head -1`
b=wc -l filename
c=`expr b - a`
tail -$c filename

?

[Victor BERRIDGE]

TOT=`cat gogo|wc -l`
let COUNT=0
for LINE in `cat gogo`
do
COUNT=`expr $COUNT + 1`
if [ "`echo $LINE|cut -c 61-62 | grep 08`" -eq 08 ]
then
TAIL=`expr $TOT - $COUNT `
tail -n`expr $TAIL` gogo>gugi
exit
fi
done

[Emilio Sierra · Accepted solution]

In the previous one there was an error:

a=`grep -n '^.{60}08' filename | cut -d':' -f1|head -1`
b=`cat filename | wc -l`
c=`expr $b - $a`
tail -$c filename

Explanation:
in $a there is the number of the first coincident line
in $b the line number of the file
in $c the subtraction between the total number and first

Excuse by my English, is poor.

[Peter Scheer]

Try:

sed -n '/^.{60}08/,$p' filename

This gives sed the address-range from (first line with 08 in cols 61,62) to ($ meaning last line) to p (print)

Regards,
Peter

[Peter Scheer]

Additional Info to previous post:
The Backslashes before the two curly braces got eaten.

Peter

[John Hall]

All above responses are close but no cigar! They do not take into account that the lines to be extracted might NOT contain 08 in columns 61 & 62. Dave said he wanted the lines AFTER the first line that met the test condition. Unfortunately Dave did not specify that these lines did or did not contain 08 (in col 61&62). If I were to cover all the bases I would have to assume Dave meant ALL lines after the line that met his criteria. The following code is more complete given the program specs Dave supplied. In my example, the input file is /tmp/abc and the output file is /tmp/xyz:

awk '{if(substr($0,61,2)=="08")
{while (getline){print $0}}
}' /tmp/abc > /tmp/xyz

Bye!

[John Hall]

All above responses are close but no cigar! They do not take into account that the lines to be extracted might NOT contain 08 in columns 61 & 62. Dave said he wanted the lines AFTER the first line that met the test condition. Unfortunately Dave did not specify that these lines did or did not contain 08 (in col 61&62). If I were to cover all the bases I would have to assume Dave meant ALL lines after the line that met his criteria. The following code is more complete given the program specs Dave supplied. In my example, the input file is /tmp/abc and the output file is /tmp/xyz:

awk '{if(substr($0,61,2)=="08")
{while (getline){print $0}}
}' /tmp/abc > /tmp/xyz

Bye!

[Dave Walley]

Thank you for you ideas, I was able to use parts from different people.

Thank you again.

Dave