Shell Programming and Scripting

hi guys i have this file with column number 7 as below:
0
1416
49
37
5
3
2
0
0
0
21
0
26
6
13
0
12
0
55
1
1
1
0
0
2
0
0
0
Device
when i select $7 i get all the values including "Device".
how do i extract only the numbers from the first 0 to the last 0 in awk?
thanks

Assuming you want to print only the numeric fields until the first non-numeric field of your example:

Code:

awk 'int($7)==0 && $7!="0"{exit}{print $7}' file

thanks that helped..

now i have this file which has the following as the last three fields..
the first fields have variable fields including spaces..

0 0.00 5
1416 85.82 1200
49 2.97 36
37 2.24 44
5 0.30 7
3 0.18 10
2 0.12 5
0 0.00 3
0 0.00 4
0 0.00 3
21 1.27 9
0 0.00 3
26 1.58 40
6 0.36 4
13 0.79 15
0 0.00 3
12 0.73 12
0 0.00 3
55 3.33 23
1 0.06 3
1 0.06 6
1 0.06 3
0 0.00 3
0 0.00 20
2 0.12 51
0 0.00 77
0 0.00 3
0 0.00 10

i want to print the column with 0,1416,49,...

i prefer to start from the last column..i know we can use $NF for last.. is there any way to use NF-3 to print the column containing 0,1416,49,...

thanks again..

Have you try it out for yourself?

Regards

ok i got it..
i am using the following:

awk '{i=NF-1;printf "\n%d", $i;}' try2 > try3

ok one last question. i have this one file and i am formatting (removing lines, extracting columns etc) right now i am creating a new file for every formatting.i have five things to extract so i generate 5 files.it is really inefficient. is there an easier way?

Perhaps you can do the job with one awk command instead of using temporary files and more awk commands.
I should start a new thread. Post the original file and the desired output within code tags.

Regards