I was trying to use bash's string replace function in xargs. But I keep getting bad substitution errors. The command I tried to use:

ls / | xargs -I{} sh -c 'x={};echo ${x/something/somethingelse}'

I could do it with sed, but I cannot figure out what's the problem with the above command.

Error message: sh: 1: Bad substitution

As KamilCuk has stated below, this was mostly due to using dash as /bin/sh.

CodePudding user response：

Issues:

• ${var/a/b} expansion is a Bash feature, which may be not available on other shells (most notably dash).

• Do not parse ls. https://unix.stackexchange.com/questions/128985/why-not-parse-ls-and-what-to-do-instead

• 'x={}; the {} part is getting interpreted by the shell, according to shell gramma rules.

• xargs by default interprets " ' \ in input strings.

Remedies:

• Call Bash explicitly if you intent to use Bash features.
• Do not use ls. (Or ignore best practice)
• Do no use {} inside -c scripts. Use arguments. Remember to quote argument expansions.
• Explicitly specify delimiter for xargs.

For example:

printf "%s\n" /* | xargs -d '\n' -n1 bash -c 'x="$1"; echo ${x/something/somethingelse}' --
printf "%s\n" /* | xargs -d '\n' -I{} bash -c 'x="$1"; echo ${x/something/somethingelse}' -- {}
find / -maxdepth 1 -mindepth 1 | xargs -d '\n' -n1 bash -c 'x="$1"; echo ${x/something/somethingelse}' --
find / -mindepth 1 -mindepth 1 -print0 | xargs -0 -n1 bash -c 'x="$1"; echo ${x/something/somethingelse}' --