Running multiple commands in one line in shell

You are using | (pipe) to direct the output of a command into another command. What you are looking for is && operator to execute the next command only if the previous one succeeded:

cp /templates/apple /templates/used && cp /templates/apple /templates/inuse && rm /templates/apple

Or

cp /templates/apple /templates/used && mv /templates/apple /templates/inuse

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To summarize (non-exhaustively) bash's command operators/separators:

• | pipes (pipelines) the standard output (stdout) of one command into the standard input of another one. Note that stderr still goes into its default destination, whatever that happen to be.
• |&pipes both stdout and stderr of one command into the standard input of another one. Very useful, available in bash version 4 and above.
• && executes the right-hand command of && only if the previous one succeeded.
• || executes the right-hand command of || only it the previous one failed.
• ; executes the right-hand command of ; always regardless whether the previous command succeeded or failed. Unless set -e was previously invoked, which causes bash to fail on an error.

Why not cp to location 1, then mv to location 2. This takes care of "removing" the original.

And no, it's not the correct syntax. | is used to "pipe" output from one program and turn it into input for the next program. What you want is ;, which seperates multiple commands.

cp file1 file2 ; cp file1 file3 ; rm file1

If you require that the individual commands MUST succeed before the next can be started, then you'd use && instead:

cp file1 file2 && cp file1 file3 && rm file1

That way, if either of the cp commands fails, the rm will not run.