Linux下外接两块触摸显示器时，不做任何处理的情况下，触摸A屏，可能会导致B屏响应，解决办法如下：
#!/bin/bash
MYID_FIRST=$1
MYID_SECOND=$2
OUT_PUT=$3
for I in $(xinput list --id-only)
do
CUR_FIRST=$(xinput list-props $I|grep "Device Product ID"|awk -F : '{print $2}'|awk -F , '{print $1}'|awk '{print $1}')
CUR_SECOND=$(xinput list-props $I|grep "Device Product ID"|awk -F : '{print $2}'|awk -F , '{print $2}'|awk '{print $1}')
if [ "$MYID_FIRST" == "$CUR_FIRST" ] && [ "$MYID_SECOND" == "$CUR_SECOND" ];
then
MY_INDEX=$I
break;
fi
done
xinput map-to-output $MY_INDEX $OUT_PUT
#xrandr 查看显示器端口（HDMI1和HDMI2参数）
#xinput 查看触摸屏id
#xinput list-props id
#export DEVICE_NAME="Multi touch Multi touch overlay device"
#./mxinput.sh 8238 6 HDMI2
#./mxinput.sh 1111 4282 HDMI1