如何在python中传递文件名作为函数参数

Question

我得到以下错误：remove_metadata((接受1个位置参数，但2个被赋予

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)