ls -U don't work?

[Alex V. Petrov]

В письме от 13 июля 2013 09:20:38 пользователь Kevin Oberman написал:


On Sat, Jul 13, 2013 at 3:30 AM, Alex V. Petrov <alexvpetrov at gmail.com[1]> wrote:


В письме от 13 июля 2013 13:45:30 Вы написали:


> 2013/7/13 Alex V. Petrov <_alexvpetrov at gmail.com_>:> > subj:> > ls -Ul>> So, what's 
the problem? Looking at code,> I'd suggest to try adding -t to actually enable sorting,> 
otherwise it just prints dates but not sorts.>> > total 0> > -rw-r--r--  1 alex  alex  0 13 июл 
15:26 000> > -rw-r--r--  1 alex  alex  0 13 июл 15:25 111> > -rw-r--r--  1 alex  alex  0 13 июл 
15:25 222> > -rw-r--r--  1 alex  alex  0 13 июл 15:25 333>> -t enables sorting (by modified 
date by default),> -U further specifies to sort by birth time.



What is the meaning of '-U'?I thought '-U' = '-tr'

man ls:"-U      Use time when file was created for sorting or printing."

-----Alex V. Petrov




No, '-U' tells ls(1) to use ctime for all timestamps. By default it uses mtime. So 'ls -U' will 
print ctime, but to sort by ctime you need 'ls -Ut' or 'ls -Utr'. 


-- R. Kevin Oberman, Network EngineerE-mail: rkoberman at gmail.com[2]

Yes. I knew already.Thank you.

-- 
-----
Alex V. Petrov


--------
[1] mailto:alexvpetrov at gmail.com
[2] mailto:rkoberman at gmail.com