Getting directory of a file in Unix

Question

I have a requirement where I need to copy some files from one location to other (Where the file may exist). While doing so,

I need to take a backup if the file already exists.
Copy the new file to the same location

I am facing problem in point 2. While I am trying to get the destination path for copying files, I am unable to extract the directory of the file. I tried using various options of find command, but was unable to crack it.
I need to trim the file name from the full file path so that it can be used in cp command. I am new to shell scripting. Any pointers are appreciated.

Diego Torres Milano · Accepted Answer · 2013-08-06 18:07:01Z

2

You can use

cp --backup

-b'--backup[=METHOD]' *Note Backup options::. Make a backup of each file that would otherwise be overwritten or removed. As a special case, `cp' makes a backup of SOURCE when the force and backup options are given and SOURCE and DEST are the same name for an existing, regular file. One useful application of this combination of options is this tiny Bourne shell script:
      #!/bin/sh
      # Usage: backup FILE...
      # Create a GNU-style backup of each listed FILE.
      for i; do
        cp --backup --force -- "$i" "$i"
      done

answered Aug 6, 2013 at 18:07

Diego Torres Milano

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2 Comments

Pramod Karandikar Over a year ago

Thanks, but when I am using -b, I am getting error that "b" is not a recognized flag.

Diego Torres Milano Over a year ago

Install GNU cp from coreutils

Mani · Accepted Answer · 2013-08-07 17:02:54Z

0

If you need only the filename, why not do a

basename /root/wkdir/index.txt

and assign it to a variable which would return only the filename?

answered Aug 7, 2013 at 17:02

Mani

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1 Comment

Pramod Karandikar Over a year ago

mani and dtmilano: Thanks for your responses. Found another way of doing it, I am now using dirname command to extract the file directory.

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