Hide

What's this?

commandlinefu.com is the place to record those command-line gems that you return to again and again.

Delete that bloated snippets file you've been using and share your personal repository with the world. That way others can gain from your CLI wisdom and you from theirs too. All commands can be commented on, discussed and voted up or down.


If you have a new feature suggestion or find a bug, please get in touch via http://commandlinefu.uservoice.com/

Get involved!

You can sign-in using OpenID credentials, or register a traditional username and password.

First-time OpenID users will be automatically assigned a username which can be changed after signing in.

Hide

Stay in the loop…

Follow the Tweets.

Every new command is wrapped in a tweet and posted to Twitter. Following the stream is a great way of staying abreast of the latest commands. For the more discerning, there are Twitter accounts for commands that get a minimum of 3 and 10 votes - that way only the great commands get tweeted.

» http://twitter.com/commandlinefu
» http://twitter.com/commandlinefu3
» http://twitter.com/commandlinefu10

Subscribe to the feeds.

Use your favourite RSS aggregator to stay in touch with the latest commands. There are feeds mirroring the 3 Twitter streams as well as for virtually every other subset (users, tags, functions,…):

Subscribe to the feed for:

Hide

News

2011-03-12 - Confoo 2011 presentation
Slides are available from the commandlinefu presentation at Confoo 2011: http://presentations.codeinthehole.com/confoo2011/
2011-01-04 - Moderation now required for new commands
To try and put and end to the spamming, new commands require moderation before they will appear on the site.
2010-12-27 - Apologies for not banning the trolls sooner
Have been away from the interwebs over Christmas. Will be more vigilant henceforth.
2010-09-24 - OAuth and pagination problems fixed
Apologies for the delay in getting Twitter's OAuth supported. Annoying pagination gremlin also fixed.
Hide

Tags

Hide

Functions

Get the date for the last Saturday of a given month

Terminal - Get the date for the last Saturday of a given month
desiredDay=6; year=2012; month=5; n=0; while [ $(date -d "$year-$((month+1))-1 - $n day" "+%u") -ne $desiredDay ]; do n=$((n+1)); done; date -d "$year-$((month+1))-1 - $n day" "+%x"
2012-05-17 12:02:30
Functions: date
0
Get the date for the last Saturday of a given month

Choosing your year and month. You only need the gnu date command and bash. desiredDay of the week is (1..7); 1 is Monday.

If you want desiredDay of week (0..6); 0 is Sunday

desiredDay=6; year=2012; month=5; n=0; while [ $(date -d "$year-$((month+1))-1 - $n day" "+%w") -ne $desiredDay ]; do n=$((n+1)); done; date -d "$year-$((month+1))-1 - $n day" "+%x"

Alternatives

There are 3 alternatives - vote for the best!

Terminal - Alternatives
cal 04 2012 | awk '{ $7 && X=$7 } END { print X }'
2012-05-06 23:43:21
User: flatcap
Functions: awk cal
2

If your locale has Monday as the first day of the week, like mine in the UK, change the two $7 into $6

cal 04 2012 | awk 'NF <= 7 { print $7 }' | grep -v "^$" | tail -1
2012-05-03 16:57:45
User: javidjamae
Functions: awk cal grep tail
-2

This is a little trickier than finding the last Sunday, because you know the last Sunday is in the first position of the last line. The trick is to use the NF less than or equal to 7 so it picks up all the lines then grep out any empty lines.

Know a better way?

If you can do better, submit your command here.

Your point of view

You must be signed in to comment.

Related sites and podcasts