问题

I'm trying to create a Bash script that will extract the last parameter given from the command line into a variable to be used elsewhere. Here's the script I'm working on:

#!/bin/bash
# compact - archive and compact file/folder(s)

eval LAST=\$$#

FILES="$@"
NAME=$LAST

# Usage - display usage if no parameters are given
if [[ -z $NAME ]]; then
  echo "compact <file> <folder>... <compressed-name>.tar.gz"
  exit
fi

# Check if an archive name has been given
if [[ -f $NAME ]]; then
  echo "File exists or you forgot to enter a filename.  Exiting."
  exit
fi

tar -czvpf "$NAME".tar.gz $FILES

Since the first parameters could be of any number, I have to find a way to extract the last parameter, (e.g. compact file.a file.b file.d files-a-b-d.tar.gz). As it is now the archive name will be included in the files to compact. Is there a way to do this?

回答1:

To remove the last item from the array you could use something like this:

#!/bin/bash

length=$(($#-1))
array=${@:1:$length}
echo $array

Even shorter way:

array=${@:1:$#-1}

But arays are a Bashism, try avoid using them :(.

回答2:

last_arg="${!#}"

回答3:

Portable and compact solutions

This is how I do in my scripts

last=${@:$#} # last parameter 
other=${*%${!#}} # all parameters except the last

EDIT
According to some comments (see below), this solution is more portable than others.
Please read Michael Dimmitt's commentary for an explanation of how it works.

回答4:

Several solutions have already been posted; however I would advise restructuring your script so that the archive name is the first parameter rather than the last. Then it's really simple, since you can use the shift builtin to remove the first parameter:

ARCHIVENAME="$1"
shift
# Now "$@" contains all of the arguments except for the first

回答5:

Thanks guys, got it done, heres the final bash script:

#!/bin/bash
# compact - archive and compress file/folder(s)

# Extract archive filename for variable
ARCHIVENAME="${!#}"

# Remove archive filename for file/folder list to backup
length=$(($#-1))
FILES=${@:1:$length} 

# Usage - display usage if no parameters are given
if [[ -z $@ ]]; then
  echo "compact <file> <folder>... <compressed-name>.tar.gz"
  exit
fi

# Tar the files, name archive after last file/folder if no name given
if [[ ! -f $ARCHIVENAME ]]; then
  tar -czvpf "$ARCHIVENAME".tar.gz $FILES; else
  tar -czvpf "$ARCHIVENAME".tar.gz "$@"
fi

回答6:

Just dropping the length variable used in Krzysztof Klimonda's solution:

(
set -- 1 2 3 4 5
echo "${@:1:($#-1)}"       # 1 2 3 4
echo "${@:(-$#):($#-1)}"   # 1 2 3 4
)

回答7:

I would add this as a comment, but don't have enough reputation and the answer got a bit longer anyway. Hope it doesn't mind.

As @func stated:

last_arg="${!#}"

How it works:

${!PARAM} indicates level of indirection. You are not referencing PARAM itself, but the value stored in PARAM ( think of PARAM as pointer to value ).
${#} expands to the number of parameters (Note: $0 - the script name - is not counted here).

Consider following execution:

$./myscript.sh p1 p2 p3

And in the myscript.sh

#!/bin/bash

echo "Number of params: ${#}"  # 3
echo "Last parameter using '\${!#}': ${!#}"  # p3
echo "Last parameter by evaluating positional value: $(eval LASTP='$'${#} ; echo $LASTP)"  # p3

Hence you can think of ${!#} as a shortcut for the above eval usage, which does exactly the approach described above - evaluates the value stored in the given parameter, here the parameter is 3 and holds the positional argument $3

Now if you want all the params except the last one, you can use substring removal ${PARAM%PATTERN} where % sign means 'remove the shortest matching pattern from the end of the string'.

Hence in our script:

echo "Every parameter except the last one: ${*%${!#}}"

You can read something in here: Parameter expansion

回答8:

#!/bin/bash

lastidx=$#
lastidx=`expr $lastidx - 1`

eval last='$'{$lastidx}
echo $last

回答9:

Are you sure this fancy script is any better than a simple alias to tar?

alias compact="tar -czvpf"

Usage is:

compact ARCHIVENAME FILES...

Where FILES can be file1 file2 or globs like *.html

回答10:

Try:

if [ "$#" -gt '0' ]; then
    /bin/echo "${!#}" "${@:1:$(($# - 1))}
fi

回答11:

Array without last parameter:

array=${@:1:$#-1}

But it's a bashism :(. Proper solutions would involve shift and adding into variable as others use.

回答12:

Alternative way to pull the last parameter out of the argument list:

eval last="\$$#"
eval set -- `awk 'BEGIN{for(i=1;i<'$#';i++) printf " \"$%d\"",i;}'`

回答13:

#!/bin/sh

eval last='$'$#
while test $# -gt 1; do
    list="$list $1"
    shift
done

echo $list $last

回答14:

I can't find a way to use array-subscript notation on $@, so this is the best I can do:

#!/bin/bash

args=("$@")
echo "${args[$(($#-1))]}"

回答15:

This script may work for you - it returns a subrange of the arguments, and can be called from another script.

Examples of it running:

$ args_get_range 2 -2 y a b "c 1" d e f g                          
'b' 'c 1' 'd' 'e'

$ args_get_range 1 2 n arg1 arg2                                   
arg1 arg2

$ args_get_range 2 -2 y arg1 arg2 arg3 "arg 4" arg5                
'arg2' 'arg3'

$ args_get_range 2 -1 y arg1 arg2 arg3 "arg 4" arg5                
'arg2' 'arg3' 'arg 4'

# You could use this in another script of course 
# by calling it like so, which puts all
# args except the last one into a new variable
# called NEW_ARGS

NEW_ARGS=$(args_get_range 1 -1 y "$@")

args_get_range.sh

#!/usr/bin/env bash

function show_help()
{
  IT="
  Extracts a range of arguments from passed in args
  and returns them quoted or not quoted.

  usage: START END QUOTED ARG1 {ARG2} ...

  e.g. 

  # extract args 2-3 
  $ args_get_range.sh 2 3 n arg1 arg2 arg3
  arg2 arg3

  # extract all args from 2 to one before the last argument 
  $ args_get_range.sh 2 -1 n arg1 arg2 arg3 arg4 arg5
  arg2 arg3 arg4

  # extract all args from 2 to 3, quoting them in the response
  $ args_get_range.sh 2 3 y arg1 arg2 arg3 arg4 arg5
  'arg2' 'arg3'

  # You could use this in another script of course 
  # by calling it like so, which puts all
  # args except the last one into a new variable
  # called NEW_ARGS

  NEW_ARGS=\$(args_get_range.sh 1 -1 \"\$@\")

  "
  echo "$IT"
  exit
}

if [ "$1" == "help" ]
then
  show_help
fi
if [ $# -lt 3 ]
then
  show_help
fi

START=$1
END=$2
QUOTED=$3
shift;
shift;
shift;

if [ $# -eq 0 ]
then
  echo "Please supply a folder name"
  exit;
fi

# If end is a negative, it means relative
# to the last argument.
if [ $END -lt 0 ]
then
  END=$(($#+$END))
fi

ARGS=""

COUNT=$(($START-1))
for i in "${@:$START}"
do
  COUNT=$((COUNT+1))

  if [ "$QUOTED" == "y" ]
  then
    ARGS="$ARGS '$i'"
  else
    ARGS="$ARGS $i"
  fi

  if [ $COUNT -eq $END ]
  then
    echo $ARGS
    exit;
  fi
done
echo $ARGS

来源：https://stackoverflow.com/questions/1215538/extract-parameters-before-last-parameter-in

标签

bash

parameters

Extract parameters before last parameter in “$@”

问题