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I'd like to construct a Linux command to list all files (with their full paths) within a specific directory (and subdirectories) ordered by access time.
ls can order by access time, but doesn't give the full path. find gives the full path, but the only control you have over the access time is to specify a range with -atime N (accessed at least 24*N hours ago), which isn't what I want.
Is there a way to order by access time and get the full path at once? I could just write a script, but it seems there should be a way to do this with the standard Linux programs.
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Listing the full path The command DIR /b will return just a list of filenames, when displaying subfolders with DIR /b /s the command will return a full pathname. To list the full path without including subfolders, use the WHERE command.
The dir command displays a list of files and subdirectories in a directory. With the /S option, it recurses subdirectories and lists their contents as well.
The ls command is used to list files or directories in Linux and other Unix-based operating systems. Just like you navigate in your File explorer or Finder with a GUI, the ls command allows you to list all files or directories in the current directory by default, and further interact with them via the command line.
ls command ls – Listing contents of directory, this utility can list the files and directories and can even list all the status information about them including: date and time of modification or access, permissions, size, owner, group etc.
find . -type f -exec ls -l {} \; 2> /dev/null | sort -t' ' -k +6,6 -k +7,7 This will find all files, and sort them by date and then time. You can then use awk or cut to extract the dates and files name from the ls -l output
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you could try:
ls -l $(find /foo/bar -type f ) ls command to achieve your goal.find cmd
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