3

I need to have an "image" of a file system without actual contents of the files - just all the names and the structure, so that I can read the file and know what files were stored there and how were they located. As always in these kinds of cases, I tend to believe that there is a beautiful "Unix way" to achieve this with a combination of some standard GNU command-line utilities. Am I right? What is it?

  • I've added find tag as the command seems to be the solution. – Ivan Dec 31 '11 at 2:30
6

Like find / -type f > /tmp/list_of_all_the_files.txt ?

  • Cool! I thought it was even going to be not a command but a script, and you come up with such a simple perfect solution! – Ivan Dec 31 '11 at 2:27
3

You can run

ls -R | grep ':$' | sed -e 's/:$//' -e 's/[^-][^\/]*\//--/g' -e 's/^/   /' -e 's/-/|/'

It shows all subdirectories niftly formatted as a tree.

Sometimes there is also tree utility available in many linux distributions. If not, then you can install it from here.

It provides following output:

~> tree -d /proc/self/
/proc/self/
|-- attr
|-- cwd -> /proc
|-- fd
|   `-- 3 -> /proc/15589/fd
|-- fdinfo
|-- net
|   |-- dev_snmp6
|   |-- netfilter
|   |-- rpc
|   |   |-- auth.rpcsec.context
|   |   |-- auth.rpcsec.init
|   |   |-- auth.unix.gid
|   |   |-- auth.unix.ip
|   |   |-- nfs4.idtoname
|   |   |-- nfs4.nametoid
|   |   |-- nfsd.export
|   |   `-- nfsd.fh
|   `-- stat
|-- root -> /
`-- task
    `-- 15589
        |-- attr
        |-- cwd -> /proc
        |-- fd
        | `-- 3 -> /proc/15589/task/15589/fd
        |-- fdinfo
        `-- root -> /

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