Question

我正在为我的基于文本的RPG改造我的战斗系统，其中一个核心内容正在丢失。我有这样的代码：

gamemap=[["player", "", "", "", "", ""], 
["", "", "", "", "", ""], 
["", "", "", "", "", ""], 
["", "", "", "", "", ""], 
["", "", "", "", "", ""], 
["", "", "", "", "", "enemy"]]


while True:
    print("╔▒▒▒▒▒▒▒▒▒▒▒╗")
    for i in gamemap:
        print("▒║", end=" ")
        for o in i:
            if o=="player":
                print("⌂", end="")
            if o=="enemy":
                print("⦿", end="")
            if o=="":
                print(".", end="")
        print("  ║▒ ")

    print("╚▒▒▒▒▒▒▒▒▒▒▒╝")


    #shows possible actions

    print("")
    print("    UP       |     ATTACK")
    print("LEFT   RIGHT | DEFEND   SPELL")
    print("    DOWN     |     QUICK")

    #asks what player wants to do
    act=str(input("Type here. "))

    if act=="RIGHT":
        ...

如果您运行此程序，它将显示我的游戏的样子。那个像房子一样的角色是玩家，而宝珠则是敌人。这是我的问题：

A）我希望那个球员可以移动。因此，如果玩家输入RIGHT，它将向右移动一个点。我怎么可能那样做？

B）如果你能够/想要你可以回答的其他事情。我想检测玩家和敌人之间的距离到int变量“距离”。以后它会对战斗系统很方便。

Answer 1

您的游戏地图是网格（或矩阵）。尝试将游戏逻辑和渲染逻辑相互分离。我的意思是;给你的玩家和敌人更多信息。一种简单的方法如下：

class Entity:
    def __init__(self, x, y, char, name):
        self.pos = (x, y)
        self.char = char
        self.name = name
    def __str__(self):
        return self.char
    def up(self):
        y = max(self.pos[1] - 1, 0)
        self.pos = (self.pos[0], y)
    def right(self, width):
        # not as nice, as up, as the width has to be
        # known for the boundaries.
        x = (self.pos[0] + 1) % width
        self.pos = (x, self.pos[1])
    # make something like this for every direction. problem:
    #  needs size of matrix for outer-bounds

player = Entity(0, 0, "⌂", "player")
enemy = Entitiy(5, 5, "⦿", "enemy")

def render(entities, width, height):
    # just basics, no borders etc. for now
    matrix = [['.' for x in range(width)] for y in range(height)]
    for entity in entities:
        x, y = entity.pos
        matrix[y][x] = str(entity)   # or entity.char, if unclear
    printable = '\n'.join(''.join(line) for line in matrix)
    print(printable)

# call with
>>> render([player, enemy], 6, 6)
⌂.....
......
......
......
......
.....⦿
>>> player.down()
......    
⌂.....
......
......
......
.....⦿

这不是最佳解决方案，因为实体是硬连线的。我建议使用Matrix - 类，保留这些结构并管理打印矩阵。

对于您的第二个问题：与上述解决方案一样，您可以将实体放置在坐标系中，可以采用不同的方法来获得距离。

<强> Euklidian geometry

import math
a = max(player.pos[0]-enemy.pos[0], enemy.pos[0]-player.pos[0]) # on x-axis (dunno, if abs would do the same)
b = max(player.pos[1]-enemy.pos[1], enemy.pos[1]-player.pos[1]) # on y-axis
distance = math.sqrt(a**2 + b**2)

<强> Manhattan-metric

aX, aY = player.pos
bX, bY = enemy.pos
distance = abs(aX - bX) + abs(aY - bY)

如果我弄错了或者你有疑问：请继续告诉我：P

修改

添加了Entity.right来说明垂直移动和矩阵边界。

Python中2D地图的运动

1 个答案: