如何解决代码“ for”,“ in”,“ not in”不起作用,如何解决,使其尊重命令之前给出的答案?
我一直在尝试制作一个简单的游戏,您可以在其中行走,并且找到不同的东西,但是由于某些原因,当我在命令中键入for和in时,它不尊重b中的答案
#
print('You may now begin your quest because I said so')
print('You must choose a number between 1 and 10,each number must be selected only once')
a = int(input('How many block do you wanna walk?'))
if a == 1:
print('Really? 1 block? LUL')
if a == 2:
b = str(input("You've found a hole,it is too dark too see what's on it though,do you wish to explore?"))
hang = ("yes")
for b in hang:
print('You fell and died.')
else:
print('Ok then,keep going.')
如您所见,它非常简单,但始终同时显示for b in hang:
和else
我确实尝试创建另一个列表fk = ("no")
并更改else
for for b in fk: print('Ok then,keep going')
,但是它也不起作用,我也尝试放置for b not in hang: print('Ok then,keep going')
,但是梯子根本不起作用
解决方法
仅通过阅读代码,您就好像在迭代字母$userIds = User::all()->pluck('id')->toArray();
//grab all posts from all users. list by most recent in descending order store in recentPosts variable
$recentPosts = Post::whereIn('user_id',$userIds)->paginate(32);
,y
和e
您到底想做什么?
如果s
应该是一个元组,请尝试在其中添加一个hang
:
,
,
您可以尝试:
def check_account(cur,account_id):
sql = """SELECT description FROM accounts WHERE id = %s;"""
acc_desc = None
try:
cur.execute(sql,(account_id,))
acc_desc = cur.fetchone()[0]
except (Exception,psycopg2.DatabaseError) as error:
print(error)
return acc_desc
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