sed - Getting Day Of Week in bash script -


i want have day of week in variable dow.

so use following bash-script:

dom=$(date +%d) dow=($($dom % 7) ) | sed 's/^0*//'

but there message bash: 09: command not found. wished result 2 ( 9 % 7 = 2) in variable $dow.

how realize this? code works days 1-8 of c-hex there no number on 8 available , message bash: 09: value great base (error token "09") appears.

you can use - flag:

dom=$(date +%-d)              ^

which prevent day being padded 0.

from man date:

   -      (hyphen) not pad field

observe difference:

$ dom=$(date +%d) $ echo $((dom % 7)) bash: 09: value great base (error token "09") $ dom=$(date +%-d) $ echo $((dom % 7)) 2

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