Bash – Using | pipe character from a $variable makes it treat as just another argument in bash; how to escape it

bashpipequotingshellvariable

I have a bash script like this

export pipedargument="| sort -n"
ls $pipedargument

But it gives the error

ls: |: No such file or directory
ls: sort: No such file or directory

It seems to be treating the contents of "| sort -n" as just an argument passed to ls.

How can I escape it so that it's treated as a regular piped command?

I'm trying to conditionally set the $pipedargument. I guess I could just conditionally execute different versions of the command but still wondering if there's a way to make this work like above?

Best Answer

You are right that you cannot use | that way. The reason is that the shell has already looked for pipelines and separated them into commands before it does the variable substitution. Hence, | is treated as just another character.

One possible work-around is to place the pipe character literally:

$ cmd="sort -n"
$ ls | $cmd

In the case that you don't want a pipeline, you can use cat as a "nop" or placeholder:

$ cmd=cat
$ ls | $cmd

This method avoids the subtleties of eval. See also here.

A better approach: arrays

A more sophisticated approach would use bash arrays in place of plain strings:

$ cmd=(sort -n)
$ ls | "${cmd[@]}"

The advantage of arrays becomes important as soon as you need the command cmd to contain quoted arguments.

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Bash – shell script options pass through to sub-command

If you specify that -- marks the end of options and all following arguments will be passed through to the subcommand, you can do something like you might find in man getopts:

       aflag=
       bflag=
       while getopts ab: name
       do
           case $name in
           a)    aflag=1;;
           b)    bflag=1
                 bval="$OPTARG";;
           ?)   printf "Usage: %s: [-a] [-b value] args\n" $0
                 exit 2;;
           esac
       done
       if [ ! -z "$aflag" ]; then
           printf "Option -a specified\n"
       fi
       if [ ! -z "$bflag" ]; then
           printf 'Option -b "%s" specified\n' "$bval"
       fi
       shift $(($OPTIND - 1))
       printf "Remaining arguments are: %s\n$*"

Specifically I'm referring to the very end there - getopts stops processing options when it encounters -- so all of those arguments will remain in $@. In the example above all of getopts processed arguments are shifted away and the remaining are printed out all at once as $*. If you handle it similarly, you could make the following work:

/mysqldumpwrapper.sh \
    -u username \
    -p password \
    -h localhost \
    -- \
    -now --all of these --are passed through

And to call the wrapped application:

mysqldump "$@" \
    --host=$MYSQL_HOST \
    --user=$MYSQL_USER \
    --password=$MYSQL_PASS "$DB" \
> "$FILE_DEST"

Bash – Left and right square brackets treated differently by sed/bash

[ introduces a character class and ] closes it.

If sed sees only ] (i.e. the closing command), it will be OK, sed assumes this is not special command.

Using only [ (without a closing ]) confuses sed, so it must be escaped.

On a side note, there is no need for

g (global) flag (only one substitution per line)
escape [ in replacement.

I managed your change with

sed -e "s/]/']/" -e "s/\[/['/"

Best Answer

A better approach: arrays

Related Solutions

Bash – shell script options pass through to sub-command

Bash – Left and right square brackets treated differently by sed/bash

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