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Open the Door


phoenix_nf

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i think this is the action script for the safe code:

// Action script...

on (press)
{
   if (_root.temp_item == "lkey")
   {
       if (_root.cinema.dial_symbol.digit0000.myNum == _root.num1 && _root.cinema.dial_symbol.digit000.myNum == _root.num2 && _root.cinema.dial_symbol.digit00.myNum == _root.num3 && _root.cinema.dial_symbol.digit0.myNum == _root.num4)
       {
           _root.comment.msg.gotoAndPlay("safe_open");
           _root.flg_lkey = 2;
           _root.item_base.disable(_root.item_base.lkey);
           _root.cinema.gotoAndStop("safe_open");
       }
       else
       {
           _root.comment.msg.gotoAndPlay("badnumber");
       } // end if
   }
   else if (_root.temp_item == "skey1" | _root.temp_item == "skey2")
   {
       _root.comment.msg.gotoAndPlay("badkey");
   }
   else
   {
       _root.comment.msg.gotoAndPlay("nothing_happen");
   } // end if
}

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in case you dont want to bother taking a screenshot and rotating in some image editor.

the url from the paper is:

http://takagism.fasco-cs.com/yellow_chamber.html

Today my code is 7397.

and here is the code from that movie:

// Action script...

// [Action in Frame 1]
function digit(who)
{
   switch (who)
   {
       case one:
       {
           _root.one.digit.gotoAndStop(num1 + 1);
           break;
       }
       case two:
       {
           _root.two.digit.gotoAndStop(num2 + 1);
           break;
       }
       case three:
       {
           _root.three.digit.gotoAndStop(num3 + 1);
           break;
       }
       case four:
       {
           _root.four.digit.gotoAndStop(num4 + 1);
           break;
       }
   } // End of switch
} // End of the function
_focusrect = False;
getURL("FSCommand:showmenu", 0);
getURL("FSCommand:allowscale", 0);
number_SO = SharedObject.getLocal("datas", "/");
if (number_SO.data.num1 != undefined)
{
   num1 = number_SO.data.num1;
}
else
{
   num1 = Math.floor(Math.random() * 10);
   number_SO.data.num1 = num1;
} // end if
if (number_SO.data.num2 != undefined)
{
   num2 = number_SO.data.num2;
}
else
{
   num2 = Math.floor(Math.random() * 10);
   number_SO.data.num2 = num2;
} // end if
if (number_SO.data.num3 != undefined)
{
   num3 = number_SO.data.num3;
}
else
{
   num3 = Math.floor(Math.random() * 10);
   number_SO.data.num3 = num3;
} // end if
if (number_SO.data.num4 != undefined)
{
   num4 = number_SO.data.num4;
}
else
{
   num4 = Math.floor(Math.random() * 10);
   number_SO.data.num4 = num4;
} // end if

// [Action in Frame 2]
stop();

// [Action in Frame 12]
stop();

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I'm glad I went straight to the puzzle rather than reading the thread first. Glad it worked for me. Glad I solved it. :w00t: Outside of giving the clue of persistance, I think you should not give anything else. Wouldn't want to spoil it for anyone else that comes through. The game is not beyond most people's capabilities. I don't understand the author's comment of only .1% making it through other than people must be getting much to impatient. It's not hard, just persist!

DL

EDIT> Nice find phoenix_nf.

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  • 2 months later...

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