如何在 python 中将文件名作为 function 参数传递？

Question

我收到以下错误：remove_metadata() 接受 1 个位置参数，但给出了 2 个

class Chess:
    def remove_metadata(filepath):
        with open(filepath, "r", errors="ignore") as f:
            contents = f.read()
            print(contents)
            
def main():
    c1 = Chess()
    c1.remove_metadata('chess_game.pgn')
    
if __name__ == "__main__":
    main()

Answer 1

将另一个参数self添加到remove_metadata

def remove_metadata(self,filepath):
    with open(filepath, "r", errors="ignore") as f:
        contents = f.read()
        print(contents)