获取bash脚本文件所在的目录

Question

当bash脚本文件包含在另一个脚本文件中(这使得它不同于this question)时，如何获得bash脚本文件所在的目录？

/script1.sh

. /dir2/script2.sh

/dir2/script2.sh

# echoes "/dir2"
echo whatevergetsthatdir

这就是我想要“修复”的剧本

/etc/init.d/silvercar-gameserver (每个实例都是唯一的)

#!/bin/bash
#
#       /etc/rc.d/init.d/silvercar-gameserver
#
#       Initscript for silvercar game server
#
# chkconfig: 2345 20 80
# description: lalalalala

#CONFIG
BIN=/opt/silvercar-gameserver # Want to get rid of this
CONF=/etc/opt/silvercar-gameserver

. /etc/init.d/functions
. $BIN/gameserver.sh.inc

exit 0

/opt/silvercar-gameserver/gameserver.sh.inc (不对每个安装进行更改)。在svn)

# Meant to be included from a script in init.d
# Required input:
#  CONF (e.g. /etc/opt/silvercarserver)


# -- Installation config (must provide JSVC, JAVA_HOME)
. $BIN/localconf.sh

# -- Instance config (must provide ASUSER, ASWORK)
. $CONF/conf.sh

PIDFILE=$ASWORK/running.pid
LOGDIR=$ASWORK/log
CLASS=tr.silvercar.gameserver.runner.DaemonGameServer
ARGS=$CONF

start() {
    echo "Going to start Gameserver..."

    export JAVA_HOME
    cd $BIN
    $JSVC -jvm server -procname silvercar-gameserver -pidfile $PIDFILE -user $ASUSER -outfile $LOGDIR/stdout -errfile $LOGDIR/stderr \
             -cp `cat classpath` -Dlogroot=$LOGDIR $CLASS $ARGS

    echo "Gameserver started!"
}

stop() {
    echo "Going to stop Gameserver..."
    $JSVC -stop -pidfile $PIDFILE $CLASS
    echo "Gameserver stopped!"
}

case "$1" in
    start)        
        start          
        ;;
    stop)
        stop
        ;;
    restart)
        stop
        start
        ;;
    *)
        echo "Usage: /etc/init.d/silvercar-gameserver {start|stop|restart}"
        exit 1
        ;;
esac

Answer 1

为了源文件，父脚本显然知道子脚本所在的路径。将其设置为变量，然后在子脚本中检查该变量。如果它是可用的，你知道它是来源的，你可以使用这个路径，否则在你所链接的问题中使用正常的技巧。

# script1.sh
RESOURCE_DIR=/dir2
source $RESOURCE_DIR/script2.sh

# script2.sh
if [ -z "$RESOURCE_DIR"]; then
  echo $RESOURCE_DIR
else
  echo $(dirname $0)
fi

Answer 2

如果您有bash3.0或更高版本，则可以使用BASH_SOURCE来获得所需的内容，而无需依赖调用程序为您设置任何环境变量。关于一个例子，见下文：

#test.sh
source some_directory/script2.sh

#some_directory/script2.sh
echo $BASH_SOURCE
echo $(dirname $BASH_SOURCE)

产出如下：

$ ./test.sh
some_directory/script2.sh
some_directory