linux - Shell脚本问题

Question

我正在编写一个 shell 脚本，使我开发的应用程序在 Linux 上作为服务运行。当我使用 ./smpp-daemon.sh start 时，我编写的 shell 脚本完美运行。但是，当我想使用以下语法时:sh smpp-daemon.sh start 我收到以下错误:/srv/smpp/bin/smpp-daemon.sh: 26:/srv/smpp/bin/smpp-daemon.sh: 函数:未找到/srv/smpp/bin/smpp-daemon.sh: 27: local: 不在函数中。

这是 shell 脚本:

#!/bin/bash
#
# Unity SMPP Daemon script
#
# Arsene Tochemey GANDOTE

# Set this to your Java installation
JAVA_HOME=/usr/lib/jvm/java-7-oracle

appNameLo="smpp"                               # application name with the first letter in lowercase
appName="SMPP"                                 # application name
osUser="root"                                    # OS user name for the service
osGroup="root"                                  # OS group name for the service
appHome="/srv/$appNameLo"                       # home directory of the service application
osUserHome="/$osUser"                           # home directory of the service user
appLogFile="/var/log/$appNameLo.log"                    # log file for StdOut/StdErr
maxShutdownTime=15                                  # maximum number of seconds to wait for the daemon to terminate normally
pidFile="/var/run/$appNameLo.pid"                   # name of PID file (PID = process ID number)
javaCommand="java"                                  # name of the Java launcher without the path
javaExe="$JAVA_HOME/bin/$javaCommand"               # file name of the Java application launcher executable
javaArgs="-jar $appHome/smsgh-smpp.jar"            # arguments for Java launcher
javaCommandLine="$javaExe $javaArgs"                # command line to start the Java service application
javaCommandLineKeyword="smsgh-smpp.jar"              # a keyword that occurs on the command line, used to detect an already running service process and to distinguish it from others

# Makes the file $1 writable by the group $osGroup.
function makeFileWritable {
   local filename="$1"
   touch $filename || return 1
   chgrp $osGroup $filename || return 1
   chmod g+w $filename || return 1
   return 0;
}

# Returns 0 if the process with PID $1 is running.
function checkProcessIsRunning {
   local pid="$1"
   if [ -z "$pid" -o "$pid" == " " ]; then return 1; fi
   if [ ! -e /proc/$pid ]; then return 1; fi
   return 0;
}

# Returns 0 if the process with PID $1 is our Java service process.
function checkProcessIsOurService {
   local pid="$1"
   if [ "$(ps -p $pid --no-headers -o comm)" != "$javaCommand" ]; then return 1; fi
   grep -q --binary -F "$javaCommandLineKeyword" /proc/$pid/cmdline
   if [ $? -ne 0 ]; then return 1; fi
   return 0;
}

# Returns 0 when the service is running and sets the variable $pid to the PID.
function getServicePID {
   if [ ! -f $pidFile ]; then return 1; fi
   pid="$(<$pidFile)"
   checkProcessIsRunning $pid || return 1
   checkProcessIsOurService $pid || return 1
   return 0;
}

function startServiceProcess {
   cd $appHome || return 1
   rm -f $pidFile
   makeFileWritable $pidFile || return 1
   makeFileWritable $appLogFile || return 1
   cmd="nohup $javaCommandLine >>$appLogFile 2>&1 & echo \$! >$pidFile"
   su -m $osUser -s $SHELL -c "$cmd" || return 1
   sleep 0.1
   pid="$(<$pidFile)"
   if checkProcessIsRunning $pid; then :; else
      echo -ne "\n$appName start failed, see logfile."
      return 1
   fi
   return 0; 
}

function stopServiceProcess {
   kill $pid || return 1
   for ((i=0; i<maxShutdownTime*10; i++)); do
      checkProcessIsRunning $pid
      if [ $? -ne 0 ]; then
         rm -f $pidFile
         return 0
         fi
      sleep 0.1
      done
   echo -e "\n$appName did not terminate within $maxShutdownTime seconds, sending SIGKILL..."
   kill -s KILL $pid || return 1
   local killWaitTime=15
   for ((i=0; i<killWaitTime*10; i++)); do
      checkProcessIsRunning $pid
      if [ $? -ne 0 ]; then
         rm -f $pidFile
         return 0
         fi
      sleep 0.1
      done
   echo "Error: $appName could not be stopped within $maxShutdownTime+$killWaitTime seconds!"
   return 1; 
}

function startService {
   getServicePID
   if [ $? -eq 0 ]; then echo -n "$appName is already running"; RETVAL=0; return 0; fi
   echo -n "Starting $appName   "
   startServiceProcess
   if [ $? -ne 0 ]; then RETVAL=1; echo "failed"; return 1; fi
   echo "started PID=$pid"
   RETVAL=0
   return 0; 
}

function stopService {
   getServicePID
   if [ $? -ne 0 ]; then echo -n "$appName is not running"; RETVAL=0; echo ""; return 0; fi
   echo -n "Stopping $appName   "
   stopServiceProcess
   if [ $? -ne 0 ]; then RETVAL=1; echo "failed"; return 1; fi
   echo "stopped PID=$pid"
   RETVAL=0
   return 0; 
}

function checkServiceStatus {
   echo -n "Checking for $appName:   "
   if getServicePID; then
  echo "running PID=$pid"
  RETVAL=0
   else
  echo "stopped"
  RETVAL=3
   fi
   return 0; 
}

function main {
   RETVAL=0
   case "$1" in
      start)                                               # starts the Java program as a Linux service
         startService
         ;;
      stop)                                                # stops the Java program service
         stopService
         ;;
      restart)                                             # stops and restarts the service
         stopService && startService
         ;;
      status)                                              # displays the service status
         checkServiceStatus
         ;;
      *)
         echo "Usage: $0 {start|stop|restart|status}"
         exit 1
         ;;
      esac
   exit $RETVAL
}

main $1

请协助，因为我不是 shell 脚本极客。我只能绕着走。

Answer 1

问题是双重的:

1. Bash 在 Debian、Ubuntu 等平台上不是 sh，因此当您使用 sh 执行脚本时，Bashishms 将无法工作。 Debian 使用 dash 代替 sh。
2. local 关键字实际上不是标准的 POSIX sh。 Dash 认为这是一个函数调用。

简单的解决方法:不要使用 local，确保您的变量名是唯一的。 Sh 脚本不是我的强项，但据我所知，在你的情况下，只需删除有问题的 local 关键字就可以了。