A function to find the newest file in a directory

newest () { find ${1:-\.} -type f |xargs ls -lrt ; }

By: mobidyc
2010-02-04 14:52:17

3 Alternatives + Submit Alt

What Others Think

I gave my vote to this one: newest () { find ${1:-\.} -type f |xargs ls -lrt ; } But I have to ask, why write a function for this when a simple $(ls -ltr) would do? Then again, are we assuming the last file that was modified in that directory wasn't a .file? Then we should do $(ls -altr). My $0.02.
unixhome · 616 weeks and 5 days ago
I forgot to mention that this function is better for recursive use. dis you ever try an ls -lrtR ls is not not efficiently for this.
mobidyc · 612 weeks and 1 day ago
The top command only works if the list of files is small enough to fit in a single run of ls -- the purpose of xargs is to split a very large number of arguments up according to the OS's maximum supported (usually 4096 at a time or so). So, there will be a separate call to ls -lrt for every batch of 4096 files, and the file at the bottom is the newest in the last batch, but not necessarily the newest overall. The one by glennie will work (I think), because the "sort -n" waits for the entire output of find before sorting.
thetrivialstuff · 546 weeks and 6 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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