Unix Bash Alias Command

问题:

I am trying to simplify my work with the help of Alias commands in my bash shell.

Problem Statement: I want to copy different files from different directories to one single folder. The syntax i am using here is as below

cp <folder>/<file>    <path>/file.dir

Here I want to save the destination file with filename.directory for easy identification. To achieve the same, I have written the below alias.

Alias Script

cp $Folder/$fileName ~/<path>/$fileName.$Folder

OR

cp $1/$2 ~/<path>/$2.$1

Expected output,

1. cp bin/file1 ~/Desktop/personal/file1.bin
2. cp etc/file2 ~/Desktop/personal/file2.etc*

However, It's failing at parsing the source file. i.e. $Folder is not replaced with my first argument.

cp: cannot stat `/file1': No such file or directory

I am writing the above script only to reduce my command lengths. As I am not expert in the above code, seeking any expert help in resolving the issue.

回答1:

Rather than using an alias you could use a function which you define in some suitable location such as .profile or .bashrc

For example:

mycp()
{
   folder=$1
   filename=$2

   if [ $# -ne 2 ]
   then
       echo "Two parameters not entered"
       return
   fi

   if  [ -d $folder -a -r $folder/$filename ]
   then
      cp $folder/$filename ~/playpen/$filename.$folder
   else
      echo "Invalid parameter"
   fi
}

回答2:

There is no way a bash alias can use arguments as you are trying to do. However, perl based rename can probably help you here. Note that it will effectively mv the files, not cp them.

rename 's|([^/]*)/(.*)|/home/user/path/$2.$1|' */*

Limitations: You can only process the files in 1 sub-directory level.

So, below alias can work (with above limitation):

$ alias backupfiles="rename 's|([^/]*)/(.*)|/home/user/path/\$2.\$1|'"
$ backupfiles */*

You can make more sophisticated perl expression if you want to work with multi-directory-level file structure.

回答3:

YAMC: Yet another myCp

This will work approx like cp as last argument is destination dir:

myCp() { 
    local dest="${@:$#}" file
    set -- "${@:1:$[$#-1]}"
    for file in "$@"; do
        cp -i "$file" "${dest%/}/${file##*/}.${file%%/*}"
    done
}

could be used in that way:

myCp */* /path/to/destdir

To test this, you could replace cp -i by echo "cp -i".

More sophisticated "myCp"

myCp() { 
    local OPTIND=0_opt cpopt dest="" dbgcmd='' file name base
    while getopts "ialdt:" opt; do
        case $opt in 
            a | i | l ) cpopt+=$opt    ;; # Forward [i|a|l] option to cp
            d )         dbgcmd='echo'  ;; # Debug option
            t )         dest="$OPTARG" ;; # Target directory
            * )         echo "Syntax error:" 1>&2; return ;;
        esac
    done
    shift $[OPTIND-1]
    [ -z "$dest" ] && dest="${@:$#}" && set -- "${@:1:$[$#-1]}"
    for file in "$@"; do
        name="${file##*/}" base="${file%/*}"
        $dbgcmd cp ${cpopt+-}$cpopt "$file" "${dest%/}/$name.${base##*/}"
    done
}

This will take last directory in source path as extension to copy files:

There is a -d option to don't copy anything, but show result command:

myCp -dait /tmp/dedup /tmp/so/*/*
cp -ai /tmp/so/abc/lock /tmp/dedup/lock.abc
cp -ai /tmp/so/config/folder1 /tmp/dedup/folder1.config
cp -ai /tmp/so/config/folder2 /tmp/dedup/folder2.config
cp -ai /tmp/so/config/somefile1 /tmp/dedup/somefile1.config

or

myCp -dl /tmp/so/*/* /tmp/dedup/
cp -l /tmp/so/abc/lock /tmp/dedup/lock.abc
cp -l /tmp/so/config/folder1 /tmp/dedup/folder1.config
cp -l /tmp/so/config/folder2 /tmp/dedup/folder2.config
cp -l /tmp/so/config/somefile1 /tmp/dedup/somefile1.config

回答4:

A directory contains some files say ~/Documents/file1.d contains newfile.txt

joe@indiana:~/Documents$ ls -l $file
total 1
-rw-r--r--   1 joe      staff          0 May  5 11:39 newfile.txt

Add the variable 'file' in .bashrc for example my .bashrc is shown here

alias ll='ls -la'
file=~/Documents/file1.d

Now whenever you copy to '$file' it will copy to file1.d directory under ~/Documents :)