Single digit months to single digit?

[Coolmar]

How can I take the output from "date +%m" which is "09" and change it to just a "9"? I want all the months that start with a 0 (01, 01...09) to trim off the 0.

Thanks,

[James R. Ferguson]

Hi:

# date "+%1d"

Regards!

...JRF...

[Sandman!]

Try the cmd below:

# date +'%0.m'

[James R. Ferguson · Accepted solution]

Hi (again) :

Sorry, you said "month" not "day":

# date "+%2m" #...with leading space

# date "+%1m" #...without leading space

Regards!

...JRF...

[Dennis Handly]

There are some evil European locales where this may be done automatically.
for i in $(locale -a); do
echo $i ====
LANG=$i locale -k d_fmt| grep -v %d
done

For months use:
... grep -v -e %m -e %b

[Coolmar]

Ok, the 'date "+%1m"' works perfectly, however come next month....which is 10 will the one be chopped off or will it remain 10? I just want any month with a leading 0, that the 0 is chopped.

Thanks for all the responses!

[Oviwan]

Or use awk:
date "+%m" | awk '{ gsub(/^0?/,""); print $0 }'

Regards

[Pete Randall]

Works fine:

# date 10010805
Mon Oct 1 08:05:00 EDT 2007
# date "+%1m"
10


Pete

[spex]

Alternatively, let the shell evaluate the output from 'date' as an integer:

$ echo $(( $(date "+%m") ))
9

[AwadheshPandey]

date +%m 1|bc