请大神把下面这段.sh转为.bat。并且附上一些说明，谢谢

[Bobolys]

#!/bin/sh
#说明$n：参数值；$*：参数字符串；$@：带引号的参数字符串；$#参数数目，不包含参数0；$?：上一次程序返回值；$$：本进程号

export JAVA_HOME=/jdk/jdk1.7.0_80/
export PATH=$JAVA_HOME/bin:$PATH
export LANG=en_US.UTF-8
EXEPATH=/tomcat/fileupload/VideoFileUpload
cd $EXEPATH
SCREEN_LOG_FILE=/tomcat/fileupload/applogs/log.log
CP="VideoFileUpload.jar"
if [ -d ./lib ]
then
for i in ./lib/*.jar
do
CP=${CP}":"${i}
done
fi

ARGV="$@"
if [ "x$ARGV" = "x" ]; then
ARGS="help"
fi

for ARG in $@ $ARGS
do
PIDSHELL=`ps -ef|grep "memsuser"|grep "smsServer.sh"|grep -v "grep"|awk '{print $2}'|sed -n 1p`
PIDJAVA=`ps -ef|grep "memsuser"|grep "java"|grep "VideoFileUpload"|grep -v "grep"|awk '{print $2}'|sed -n 1p`


echo "启动VideoFileUpload..."
if [ "$PIDJAVA" != "" ]; then
echo " VideoFileUpload已经启动"
else
java -Xmx128M -Xms32m -Djava.library.path=./lib -classpath ${CP} com.cmbcsa.fileuplod.client.VideoFileUpload >$SCREEN_LOG_FILE 2>&1 &
echo " VideoFileUpload启动成功"
fi
done
exit 0

[Bobolys]

实在不太明白，求讲解，求教学