[SOLVED] unix cal command special character

Issue

When I try "cal | tail -6" in my unix machine, I get -

             1  2  3
 4  5  6  7  8  9 10 
11 12 13 14 15 16 17 
18 19 20 21 22 23 24
25 26 27 28 29 30

but when I try "cal | tail -6 | awk '{print $7}'", I get -

10
17
24

where is 3 going ? My requirement is basically all weekdays i.e column 2,3,4,5 & 6. But I'm getting wrong output because of the strange behavior of "cal"

Solution

Since all of your columns in each row are three characters wide, you could use this to extract the days you wish for. For example, if you wanted only the 7th day in a column, you could do the following:

cal | sed 's/^\(.\{18\}\).*$/\1/'

This command would remove the first 18 characters in the line, which are the entries for the first 6 days of the week.

To extract a particular day, such as the fourth day, you could do this:

cal | sed 's/^.\{9\}\(.\{3\}\).*$/\1/'

To remove the first day of the week and the last day, you could do this:

cal | sed -e 's/^.\{3\}//' -e 's/^\(.\{15\}\).\{3\}$/\1/'

Answered By - djhaskin987

Answer Checked By - Senaida (WPSolving Volunteer)