awk - field sepearor

Shell Programming and Scripting
1

I have a file where each line can contain differing numbers of fields

e.g

The big red company 00:01:01 ...
The next big red company 00:00:01 ...

I need to create a variable containing only the company (i.e everything to the left of the first 00), I would normally read in each line (using while) and pipe it to awk with the -F switch however this is not appropriate in this case.

Any ideas would be appreciated

2 
BEGIN {
   FS=/ [0-9]/
}
{ print $1 }
3

When I try this it only shows 'The'

4

if on Solaris try it with '/usr/xpg4/bin/awk'

5

I'm using AIX

6

ok, and this?

BEGIN {
   FS="[ ][0-9]"
}
{ print $1 }
7

Sorted ! , many thanks

8

2 more alternatives


$ awk 'FS="[0-9][0-9]:" { print $1 }' file1
The big red company 
The next big red company 



$ cut -d":" -f 1 file1 | sed 's/...$//'
The big red company
The next big red company