正在搜索连接到列表属性的属性的特定字符串？

Question

我可能用了可怕的措辞，所以这里有一个解释：我有两个类，Tile和Grid。 在图块中，我创建了基本图块格式及其位置，名称和项目。 在Grid中，我使用for循环来正确创建这些图块并添加信息并将其全部存储在“网格”中。我想允许用户在网格内搜索图块具有等于“属性”的属性的图块用户尝试搜索的内容，然后返回所有信息（x和y位置，“名称”和“项目”属性）。

global grid

letters = ["A","B","C","D","E","F","G","H","I"]

class Tile:
    def __init__(self, x, y, name=None, items="Empty"):
        self.x = x
        self.y = y
        self.contents = {
            "name": name,
            "items": items
        }

    def __repr__(self):
        return str(self.contents)

class Grid:
    def __init__(self):
        self.grid = []
        self.resetGrid()

    def resetGrid(self):
        grid = [[Tile(x, y) for x in range(9)] for y in range(9)]
        for n in range(9):
            letter = letters[n]
            for m in range(9):
                grid[n][m].contents = {"name":letter + str(m+1),"items":"Empty"}
        return grid

    def printGrid(self):
        for row in grid:
            print()
            print("-" * 44)
            for tile in row:
                name = tile.contents["name"]
                #items = tile.contents["items"]
                print(name, end=" | ")
        print()
        print("-" * 44)

    def searchGrid(self):
        grid = self.resetGrid()
        print()
        print("")
        print("Please select which square you wish to see the information about.")
        userinput = input("> ")
        for name in tile.contents["name"]:
            pring("k")

newgrid = Grid()
newgrid.printGrid()

#newgrid.resetGrid()

谢谢你的帮助。

Answer 1

只需在每个图块中搜索名称，直到找到它即可。 这是你想要的？

def searchGrid(self):
        grid = self.resetGrid()
        print()
        print("")
        print("Please select which square you wish to see the information about.")
        userinput = input("> ")
        for row in self.grid:
            for tile in row:
                 if tile.contents['name'] == userinput:
                       print(tile.contents)
                       #other things
                       break

另请注意拼写错误：

更改pring到print

在printGrid()方法printGrid() grid更改为self.grid

Answer 2

letters = ["A","B","C","D","E","F","G","H","I"]

class Tile:
    def __init__(self, x, y, name=None, items="Empty"):
        self.x = x
        self.y = y
        self.contents = {
            "name": name,
            "items": items
        }

    def __repr__(self):
        return str(self.contents)

class Grid:
    def __init__(self):
        self.grid = []
        self.resetGrid()

    def resetGrid(self):
        self.grid = [[Tile(x, y) for x in range(9)] for y in range(9)]
        for n in range(9):
            letter = letters[n]
            for m in range(9):
                self.grid[n][m].contents = {"name":letter + str(m+1),"items":"Empty"}

    def printGrid(self):
        for row in self.grid:
            print()
            print("-" * 44)
            for tile in row:
                name = tile.contents["name"]
                #items = tile.contents["items"]
                print(name, end=" | ")
        print()
        print("-" * 44)

    def searchGrid(self):
        print()
        print("")
        print("Please select which square you wish to see the information about.")
        userinput = input("> ")
        for n in range(9):
            for m in range(9):
                if userinput == self.grid[n][m].contents['name']:
                    print(self.grid[n][m].x)
                    print(self.grid[n][m].y)
                    print(self.grid[n][m].contents['name'])
                    print(self.grid[n][m].contents['items'])


newgrid = Grid()
newgrid.printGrid()
newgrid.searchGrid()