When I use this command I get an output of some numbers
cat ac.20070511 | cut -d" " -f19
Is there any way for me to display only the numbers that are greater than 1000 but not all the numbers in the ouput.
Can any one help me with this. 
what's about:
awk '$19>999 {print $19}' ac.20070511
gP
Venu, as always, there are several ways to solve a problem.
This is one of them using part of your solution:
Note there are four dots inside of single quote.
cut -d" " -f19 input_file | egrep '....'
Pressy,
Yours should be working but the problem I have is in the output i have "12345" format. How can I eliminate "" so that i can check for the condition.
Shell,
That one was good but it filters by number of digits, but if I need those greater than 15000 only then I think it might not work. Anyway thanks for that.
ok...
:rolleyes:
# sed 's/"//g' ac.20070511 | awk '$19>999 {print $19}'
gP
Venu,
The "sed" that I wrote works for what you asked for:
Now you are changing the requirement:
There is no way one can write a solution that works after the
initial requirement changes.
a bit off-topic, but....
there's no need to use sed.
awk '{gsub("\"","",$19);if ($19>15000) print}' ac.20070511