目标是一旦p1或p2获胜，就会从“循环时最终结果”（作为注释添加）中得出。我创建了一个初始为False的全局标志，但是一旦winningfunc（）返回result1或结果2等于3，标志就会变为True。我知道可能有些冗长，但是我是python或任何类型的编码的新手。
尝试制作一个简单的井字游戏。任何帮助表示赞赏：）
frame=[[" ","|"," ","|"," "],
["-","-","-","-","-"],
[" ","|"," ","|"," "],
["-","-","-","-","-"],
[" ","|"," ","|"," "]]
# Name assignment
while True:
p1_name=input("Enter player 1 name: ")
p2_name=input("Enter player 2 name: ")
if p1_name.lower()==p2_name.lower():
print("Both names entered are same, please re-enter!")
else:
break
# Notation assignment
while True:
p1_nt=input(f"{p1_name}, choose 'X' or 'O' to play with: ")
if p1_nt.lower()=="x":
p1_nt="X"
p2_nt="O"
print(f"{p1_name} will play with 'X' & {p2_name} will play with 'O'")
break
elif p1_nt.lower()=="o":
p1_nt="O"
p2_nt="X"
print(f"{p1_name} will play with 'O' & {p2_name} will play with 'X'")
break
else:
print("Entered notation is wrong, please re-enter!")
# dict
dc={"1":[4,0],"2":[4,2],"3":[4,4],"4":[2,0],"5":[2,2],"6":[2,4],"7":[0,0],"8":[0,2],"9":[0,4]}
# winning logic definition
flag=False
def winningfunc ():
def mf(v1,v2,v3):
result1=0
result2=0
for x in range(v1,v2,v3):
if frame[dc[str(x)][0]][dc[str(x)][1]]==p1_nt:
result1+=1
elif frame[dc[str(x)][0]][dc[str(x)][1]]==p2_nt:
result2+=1
else:
pass
if result1==3:
flag=True
print(f"Congratulations, {p1_name} wins! ")
elif result2==3:
flag=True
print(f"Congratulations, {p2_name} wins! ")
else:
pass
mf(1,4,1) # for c4
mf(4,7,1) # for c2
mf(7,10,1) # for c0
mf(1,10,4) # for d1
mf(3,8,2) # for d2
mf(1,8,3) # for r0
mf(2,9,3) # for r2
mf(3,10,3) # for r3
# loop for payer input
while not flag: # for final result
while True: # for p1 data filling
s1=input(f"{p1_name} its your turn: ")
if 0<=int(s1)<=9 and frame[dc[str(s1)][0]][dc[str(s1)][1]]==" ":
if p1_nt=="X":
frame[dc[str(s1)][0]][dc[str(s1)][1]]="X"
else:
frame[dc[str(s1)][0]][dc[str(s1)][1]]="O"
break
else:
print(f"Wrong entry {p1_name}, try once again!")
x1=None
for x1 in range(len(frame)):
f1="".join(frame[x1])
print(f1)
# Result check for p1
winningfunc()
while True: # for p2 data filling
s2=input(f"{p2_name} its your turn: ")
if 0<=int(s1)<=9 and frame[dc[str(s2)][0]][dc[str(s2)][1]]==" ":
if p2_nt=="X":
frame[dc[str(s2)][0]][dc[str(s2)][1]]="X"
else:
frame[dc[str(s2)][0]][dc[str(s2)][1]]="O"
break
else:
print(f"Wrong entry {p2_name}, try once again!")
x2=None
for x2 in range(5):
f2="".join(frame[x2])
print(f2)
# Result check for p2
winningfunc()
答案 0 :(得分：3)
要访问全局flag，您需要在方法开始时指定它是全局的。此外，您应该在玩家2回合之前检查flag是否为true。这是工作代码：
frame=[[" ","|"," ","|"," "],
["-","-","-","-","-"],
[" ","|"," ","|"," "],
["-","-","-","-","-"],
[" ","|"," ","|"," "]]
# Name assignment
while True:
p1_name=input("Enter player 1 name: ")
p2_name=input("Enter player 2 name: ")
if p1_name.lower()==p2_name.lower():
print("Both names entered are same, please re-enter!")
else:
break
# Notation assignment
while True:
p1_nt=input(f"{p1_name}, choose 'X' or 'O' to play with: ")
if p1_nt.lower()=="x":
p1_nt="X"
p2_nt="O"
print(f"{p1_name} will play with 'X' & {p2_name} will play with 'O'")
break
elif p1_nt.lower()=="o":
p1_nt="O"
p2_nt="X"
print(f"{p1_name} will play with 'O' & {p2_name} will play with 'X'")
break
else:
print("Entered notation is wrong, please re-enter!")
# dict
dc={"1":[4,0],"2":[4,2],"3":[4,4],"4":[2,0],"5":[2,2],"6":[2,4],"7":[0,0],"8":[0,2],"9":[0,4]}
# winning logic definition
flag=False
def winningfunc ():
global flag
def mf(v1,v2,v3):
global flag
result1=0
result2=0
for x in range(v1,v2,v3):
if frame[dc[str(x)][0]][dc[str(x)][1]]==p1_nt:
result1+=1
elif frame[dc[str(x)][0]][dc[str(x)][1]]==p2_nt:
result2+=1
else:
pass
if result1==3:
flag=True
print(f"Congratulations, {p1_name} wins! ")
elif result2==3:
flag=True
print(f"Congratulations, {p2_name} wins! ")
else:
pass
mf(1,4,1) # for c4
mf(4,7,1) # for c2
mf(7,10,1) # for c0
mf(1,10,4) # for d1
mf(3,8,2) # for d2
mf(1,8,3) # for r0
mf(2,9,3) # for r2
mf(3,10,3) # for r3
# loop for payer input
while not flag: # for final result
while True: # for p1 data filling
s1=input(f"{p1_name} its your turn: ")
if 0<=int(s1)<=9 and frame[dc[str(s1)][0]][dc[str(s1)][1]]==" ":
if p1_nt=="X":
frame[dc[str(s1)][0]][dc[str(s1)][1]]="X"
else:
frame[dc[str(s1)][0]][dc[str(s1)][1]]="O"
break
else:
print(f"Wrong entry {p1_name}, try once again!")
x1=None
for x1 in range(len(frame)):
f1="".join(frame[x1])
print(f1)
# Result check for p1
winningfunc()
if flag:
break
while True: # for p2 data filling
s2=input(f"{p2_name} its your turn: ")
if 0<=int(s1)<=9 and frame[dc[str(s2)][0]][dc[str(s2)][1]]==" ":
if p2_nt=="X":
frame[dc[str(s2)][0]][dc[str(s2)][1]]="X"
else:
frame[dc[str(s2)][0]][dc[str(s2)][1]]="O"
break
else:
print(f"Wrong entry {p2_name}, try once again!")
x2=None
for x2 in range(5):
f2="".join(frame[x2])
print(f2)
# Result check for p2
winningfunc()