function onAfterSetState_swordman_buff(obj , state , datas , isResetTimer)
{
if(!obj) return;
local SkillIndex = obj.sq_GetState();
local apd = "";
if(SkillIndex == 60)
apd = "character/swordman/appendage/ap_flowmindpowerup.nut";
if(apd != "")
{
if(CNSquirrelAppendage.sq_IsAppendAppendage(obj , "character/swordman/appendage/ap_flowmindpowerup.nut"))
{
CNSquirrelAppendage.sq_RemoveAppendage(obj , "character/swordman/appendage/ap_flowmindpowerup.nut");
}
local validTime = obj.sq_GetLevelData(SkillIndex , 0 , sq_GetSkillLevel(obj , SkillIndex));;
local Appendage = CNSquirrelAppendage.sq_AppendAppendage(obj , obj , SkillIndex , false , apd , true);
Appendage.sq_SetValidTime(validTime);
}
}
，流心狂的状态编号怎么获取呢？已经卡在最后一步了

可以不用nut吧

次级apd无法驱散这个不是很懂，要不试试onEnd？因为按道理来说，刀疤鼠驱散流心狂主 Buff 时，主 Buff 会触发 onEnd，那在onEnd里加个对次级apd的检测，检测到次级apd就可以一起移除？我看他们写的
//ap销毁结束时
function sq_AddFunctionName(appendage)
{
appendage.sq_AddFunctionName("onEnd", "onEnd_appendage_atmage_bodyeffect")
}
function onEnd_appendage_PushOut(appendage)
{
if(!appendage) {
return;
}
local parentObj = appendage.getParent();
if(!parentObj) {
appendage.setValid(false);
return;
}
if (parentObj.getState() == STATE_HOLD) {
appendage.setValid(false);
parentObj.sendStateOnlyPacket(STATE_STAND);
}
// TODO : x축 우선 체크 할것. 지금은 y축으로 내려와 버려서 연속기가 안됨
//local ao = sq_GetCNRDObjectToActiveObject(parentObj);
//if(ao && sq_IsMyControlObject(ao) && !ao.isMovablePos(ao.getXPos(),ao.getYPos())) {
//sq_SimpleMoveToNearMovablePos(ao,100); // 이동불가 지역에 있다면 이동가능지역으로 이동
//}
}
还有男法的
function onEndState_IceCrash(obj, newState)
{
if(!obj)
return;
if(newState != STATE_ICE_CRASH)
{
removeAllIceCrashAppendage(obj);
sq_RemoveParticle("Character/Mage/Particle/IceCrashIceDust.ptl",obj);
}
}
所以感觉可以在流心狂ap的onEnd里面这么写
function onEnd(appendage)
{
local obj = appendage.getParent();
if(!obj) return;
// 刀疤鼠驱散主 Buff 时，强制搜索并移除次级 APD
if(obj.sq_HasSquirrelAppendage("次级APD路径.nut"))
{
obj.sq_RemoveSquirrelAppendage("次级APD路径.nut");
}
}