18.12.2015, 02:21
Print the value right after you declare the variable.
PHP код:
new furnitureid = CreateDynamicObject(fid, HouseInfo[x][ExitX],HouseInfo[x][ExitY],HouseInfo[x][ExitZ], 0,0,0, HouseInfo[x][VirtualWorld],HouseInfo[x][Interior]);
printf("furnitureid = %d", furnitureid);