语法错误";;意外的"关于Debian的简单初始化脚本 [英] Syntax error of ";; unexpected" on simple init script for Debian
本文介绍了语法错误";;意外的"关于Debian的简单初始化脚本的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有这个初始化脚本来运行uwsgi.它有效,但仅在启动命令上有效.所有其他命令给我这个错误:
I have this init script to run uwsgi. It works, but only on the start command. All the other commands give me this error:
/etc/init.d/uwsgi: 27: /etc/init.d/uwsgi: Syntax error: ";;" unexpected
似乎我正在阅读的教程中应该有冒号,但是它告诉我删除它们?
It seems like the colons should be there from the tutorials I am reading, and yet it is telling me to remove them?
#!/bin/sh
### BEGIN INIT INFO
# Provides: uwsgi
# Required-Start: $local_fs $remote_fs $network
# Required-Stop: $local_fs $remote_fs $network
# Default-Start: 2 3 4 5
# Default-Stop: 0 1 6
# Description: This script manages uWSGI.
### END INIT INFO
DAEMON=/var/www/app/venv/bin/uwsgi
PIDFILE=/var/run/uwsgi.pid
DAEMON_ARGS="--ini /var/www/app/conf/uwsgi/app.ini --pidfile /var/run/uwsgi.pid"
. /lib/init/vars.sh
. /lib/lsb/init-functions
case "$1" in
start)
echo "Starting"
start-stop-daemon --start --quiet --pidfile $PIDFILE --exec $DAEMON --test > /dev/null \
|| return 1
start-stop-daemon --start --quiet --pidfile $PIDFILE --exec $DAEMON -- \
$DAEMON_ARGS 1> /dev/null 2>&1 \
|| return 2
esac
;;
stop)
echo "Stopping"
start-stop-daemon --stop --quiet --retry=QUIT/30/KILL/5 --pidfile $PIDFILE --name uwsgi
RETVAL="$?"
[ "$RETVAL" = 2 ] && return 2
start-stop-daemon --stop --quiet --oknodo --retry=0/30/KILL/5 --exec $DAEMON
[ "$?" = 2 ] && return 2
rm -f $PIDFILE
return "$RETVAL"
esac
;;
status)
status_of_proc "$DAEMON" "uwsgi" && exit 0 || exit $?
;;
reload)
echo "Reloading"
start-stop-daemon --stop --signal 1 --quiet --pidfile $PIDFILE --name uwsgi
return 0
;;
*)
echo "Usage: /etc/init.d/uwsgi {start|stop|status|reload|force-reload}" >&2
exit 3
;;
esac
exit 0
推荐答案
在;;
之前删除esac
关键字.末尾应该只有一个与最初的case
关键字匹配.
Get rid of the esac
keywords before the ;;
s. There should only be one at the very end to match the initial case
keyword.
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