I am currently using fgrep in my scritps to search text files that contain millions of records. I need to print out the strings that starts with '0000000000H' only.

DIRS="AA AB AC AD"
for i in $DIRS
do
fgrep '0000000000H' ${i}.txt > new_${i}.txt
done

Is there a faster way to do this so that only the lines with '0000000000H' are printed.

Seems like this takes forever to run.

Thanks

It should be faster to use an ordinary regexp that includes the '^' beginning of line matcher instead of the fixed-string anywhere in the line method, especially if the lines are long and you can always try the --mmap switch '-)

It is also possible that sed or awk would be faster but you'd have to experiment and the simplicity of the task means that it would be easy to write a c programme which may be worth doing if it's something you'll need to run regularly or often.

sorry -- it wasn't my grep command it seems the hold up is when I am trying to remove leading and trailing white spaces.

gawk '{printf ("%s|%s|%s|%s|%s\n",substr($0,16,10),substr($0,26,14),substr($0,50,17),substr($0,56,16),substr($0,112,78))}' <test.txt | sed 's/ //g' - | sed 's/ |/|/g' - | sed 's/| /|/g' > new.txt

there is no set tab of white spaces and I do not want to put a "|" in between words in a field.

I thought that maybe using gsub would do it -- but I can't seem to get that to work

anyway of reducing the time in removing leading and trailing spaces in a field.

Thanks

think I got it

if I use gawk '{gsub( / +/, "");print }' file > newfile

this seems to work
thanks