I'm trying to open a file with vim from the command line, the file is in a directory filled with automatically generated files that are prepended with a time stamp. Since I don't know the time stamps off the top of my head, I would just like to open the most recent one in the directory (also the last one in the list alphabetically).
vim ./my_dir/<last_item>
Is there a way to do this?
This is really a job for zsh.
vim my_dir/*(om[1])
The bits in parentheses are glob qualifiers. The * before is a regular glob pattern, for example if you wanted to consider only log files you could use *.log. The o glob qualifier changes the order in which the matches are sorted; om means sort by modification time, most recent first. The [ glob qualifier means to only return some of the matches: [1] returns the first match, [2,4] return the next three, [-2,-1] return the last two and so on. If the files have names that begin with their timestamp, *([1]) will suffice.
In other shells, there's no good way to pick the most recent file. If your file names don't contain unprintable characters or newlines, you can use
vim "$(ls -t mydir | head -n 1)"
If you want to pick the first or last file based on the name, there is a fully reliable and portable method, which is a bit verbose for the command line but perfectly serviceable in scripts.
set -- mydir/*
first_file_name=$1
eval "last_file_name=\${$#}"
If you are inside my_dir:
vim $(ls -rt | tail -1)
vim "$(ls -t | head -1)"
– enzotib
Dec 09 '11 at 15:39
This works for me:
$ ls -rt | tail -1 | xargs vim
But you may have a directory as the last entry, in which case you might want to filter directories:
$ ls -rt | grep -v "^d" | tail -1 | xargs vim
I guess you are looking for this.
[centos@centos temp]$ for i in `seq 100`; do touch $i; done
[centos@centos temp]$ ls -t1 | head -1
100
[centos@centos temp]$ vim /.my_dir/$(ls -t1 | head -1)
But sorting depends entirely on your filename.
