永恒狂刀两次放入骰子验证脚本

永恒狂刀两次放入骰子验证脚本示图：

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对话NPC("元宝回收")
等待(1500)
::A::
aa=NPC对话内容
aa=string.gsub(aa,"┐","BB")
aa=string.gsub(aa,"└","BB")
aa=string.gsub(aa,"┊","")
aa=string.gsub(aa,"　","")
aa=string.gsub(aa,"●","AA")
aa=string.gsub(aa,"/SCOLOR=250","")
aa=string.gsub(aa,"<","")
aa=string.gsub(aa,">","")
aa=string.gsub(aa,"\\","")
aa=string.gsub(aa,"清理验证骰子/@清理","")
pp=aa
k1="BBAABB"
k2="BBAAAABB"
k3="BBAAAAAABB"
k4="BBAAAAAAAABB"
k5="BBAAAAAAAAAABB"
k6="BBAAAAAAAAAAAABB"
if string.find(aa,k1,1,true) then
k1=取背包物品属性("1点",21)
投递物品(tonumber(k1),"1点")
end
if string.find(aa,k2,1,true) then
k1=取背包物品属性("2点",21)
投递物品(tonumber(k1),"2点")
end
if string.find(aa,k3,1,true) then
k1=取背包物品属性("3点",21)
投递物品(tonumber(k1),"3点")
end
if string.find(aa,k4,1,true) then
k1=取背包物品属性("4点",21)
投递物品(tonumber(k1),"4点")
end
if string.find(aa,k5,1,true) then
k1=取背包物品属性("5点",21)
投递物品(tonumber(k1),"5点")
end
if string.find(aa,k6,1,true) then
k1=取背包物品属性("6点",21)
投递物品(tonumber(k1),"6点")
end
k2=string.find(pp,"投入正确的验证骰子",1,true)
if k2~=nil then
goto A
end