Shell Programming and Scripting

I have a file such as:

.....12345......67890...xxx
....123456....78901...yyy
...1234567...89012...zzz
..12345678.90123...aaa

Where the '.' character is a SPACE.

I'm trying to print just the first number in each line.

such as:

12345
123456
1234567
12345678

Both the number of spaces before the number of interest and the number of spaces after the number of interest are variable.

I have tried several cut -d " " -f(x) file >outfile options with no success.

I'd image this would be done with an 'if then' regular expression lookup.

Such as reading from left if character is a digit print character if not don't. If character in (position -1) was a digit and current character is a space then go to next line.

... or some other command I simply don't know likely exists.


Thank you.

Next time try awk

Code:

awk -F' ' '{print $1}' file