Spawn a background process in a bash function

问题

I am working on writing a Bash function to start a server that needs to be started from a certain folder, but I don't want starting this server to impact my current work. I've written the following:

function startsrv {
        pushd .
        cd ${TRUNK}
        ${SERVERCOMMAND} & 
        popd
}

My variables are all set, but when this executes, I get an error regarding an unexpected semicolon in the output, it appears that Bash is inserting a semicolon after the ampersand starting ${SERVERCOMMAND} in the background.

Is there anything I can do to start ${SERVERCOMMAND} in the background while still using pushd and popd to make sure I end up back in my current directory?

Edit: Output of echo ${SERVERCOMMAND}, since it was requested:

yeti --server --port 8727

Error message:

-bash: syntax error near unexpected token `;'

回答1:

What is the value of $SERVERCOMMAND? You must have a semi-colon in it.

For what it's worth you can simplify the pushd/cd to one pushd:

pushd $TRUNK
$SERVERCOMMAND &
popd

Or create a subshell so the cd only affects the one command:

(cd $TRUNK; $SERVERCOMMAND &)

回答2:

You can also use cd -

cd $TRUNK
$SERVERCOMMAND &
cd -

回答3:

A semicolon in ${SERVERCOMMAND} should not trigger a syntax error unless there is a bug in bash itself. The semicolon problem must be located elsewhere, in a part of the code we don't see.

Aside from the semicolon issue, there are some minor bugs in your code:

The ${TRUNK} variable expansion isn't quoted. If the directory name contains whitespace, bash will split it into multiple fields before invoking cd.
The return value of cd ${TRUNK} isn't checked. If the directory doesn't exist, bash will invoke the server in the current directory.
The function doesn't test whether ${SERVERCOMMAND} might fail to execute (e.g., command not found).
The function keyword and the pushd and popd commands are bash-specific, so this code won't run in a POSIX shell.

Here's a safer, POSIX-compatible rewrite:

log() { printf '%s\n' "$*"; }
error() { log "ERROR: $*" >&2; }
fatal() { error "$*"; exit 1; }

startsrv() {
    (
        cd "${TRUNK}" || fatal "failed to cd to '${TRUNK}'"
        set -- ${SERVERCOMMAND}
        command -v "$1" >/dev/null || fatal "command '$1' not found"
        command "$@" &
    ) || exit 1
}

来源：https://stackoverflow.com/questions/6298623/spawn-a-background-process-in-a-bash-function

标签

bash

shell

unix

bash-function