Bash ls and command pipe

bash shell unix command-line-arguments ls

You can't pipe stuff into ls.. You could do something like:

ls -ld $(cut -f 6 -d ':' /etc/passwd | sort -su)

By spawning a new bash to execute the cut | sort and passing it as a ls argument

bash shell unix command-line-arguments ls

ls does not take piped output. You could, however, use forward quotes to execute it on a list of directories:

ls `cut -f 6 -d ':' /etc/passwd | sort -su `

bash shell unix command-line-arguments ls

You could use xargs

cut -f 6 -d ':' /etc/passwd | sort -su | xargs ls

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