Printing only the first field in a string

Question

I have a date as 12/12/2013 14:32 I want to convert it into only 12/12/2013. The string can be 1/1/2013 12:32 or 1/10/2013 23:41 I need only the date part.

Answer 1

You can do this easily with a variety of Unix tools:

$ cut -d' ' -f1  <<< "12/12/2013 14:32"
12/12/2013

$ awk '{print $1}' <<< "12/12/2013 14:32"
12/12/2013

$ sed 's/ .*//' <<< "12/12/2013 14:32"
12/12/2013

$ grep -o "^\S\+"  <<< "12/12/2013 14:32"
12/12/2013

$ perl -lane 'print $F[0]' <<< "12/12/2013 14:32"
12/12/2013

Answer 2

$ echo "12/12/2013 14:32" | awk '{print $1}'
12/12/2013

print $1 --> Prints first column of the supplied string. 12/12/2013

print $2 --> Prints second column of the supplied string. 14:32

By default, awk treats the space character as the delimiter.

Answer 3

If your date string is stored in a variable, then you don't need to run an external program like cut, awk or sed, because modern shells like bash can perform string manipulation directly which is more efficient.

For example, in bash:

$ s="1/10/2013 23:41"
$ echo "${s% *}"
1/10/2013