我想检查输入的密码是否与数据库中存储的密码相同，但是当我使用它时，bcrypt.checkpw()它会返回一个错误，指出它需要一个字符串或字节，因为 SQL 查询返回一个元组。我找不到将数据库响应转换为元组中的字节以使其兼容的方法。sql = ''' SELECT password FROM user_data WHERE username=? '''username = input('Input username: ')password = bytes(input('Input Password: '), encoding='utf-8')cur = conn.cursor()cur.execute(sql, (username,))rows = cur.fetchall()for row in rows: if bcrypt.checkpw(password, row): details = (user_id, username, password) print('logged in') return details break
4 回答
德玛西亚99
TA贡献1770条经验 获得超3个赞
def compress(data):
result = []
keys = tuple(data[0].keys()) # the keys
result.append(keys)
result2 = []
result2.append(tuple([data[0][key] for key in keys]))
result2.append(tuple([data[1][key] for key in keys[::-1]]))
result.append(result2)
return tuple(result)
data = [
{"a": 1, "b": 2, "c": 3},
{"a": 4, "c": 6, "b": 5}
]
print(compress(data))
印刷：
(('a', 'b', 'c'), [(1, 2, 3), (6, 5, 4)])
湖上湖
TA贡献2003条经验 获得超2个赞
应该
def compress(data):
res = []
for idx, sub in enumerate(data, start=0):
if idx == 0:
res.append(tuple(sub.keys()))
res.append([])
res[-1].append(tuple(sub.values()))
else:
res[-1].append(tuple(sub.values()))
return tuple(res)
HUX布斯
TA贡献1876条经验 获得超6个赞
所以我创建了你想要的东西，你可以把它放在一个函数中，如果你愿意的话
data = [
{"a": 1, "b": 2, "c": 3},
{"a": 4, "c": 6, "b": 5}
]
keys = [[],[]]
for d in data:
items = list(d.items())
l = []
for i in items:
l.append(i[1])
if i[0] not in keys[0]:
keys[0].append(i[0])
else:
continue
keys[1].append(tuple(l))
keys = tuple(keys)
肥皂起泡泡
TA贡献1829条经验 获得超6个赞
使用列表理解：
def compress(data):
keys = tuple(sorted(data[0].keys()))
values = [tuple(d[k] for k in keys) for d in data]
return (keys, values)
>>> compress([{"a": 1, "b": 2, "c": 3},{"a": 4, "c": 6, "b": 5}])
(('a', 'b', 'c'), [(1, 2, 3), (4, 5, 6)])
可选：缺少钥匙
如果某些字典中可能缺少某些键，您可以使用所有字典中的所有键，然后使用 删除重复的键set，然后使用d.get(k, default_value)代替d[k]：
def compress(data, default_value=None):
keys = tuple(sorted(set(k for d in data for k in d.keys())))
values = [tuple(d.get(k, default_value) for k in keys) for d in data]
return (keys, values)
>>> data = [{'a':1, 'b': 2, 'c': 3}, {'a':11, 'b':12, 'd':14}]
>>> compress(data, 0)
(('a', 'b', 'c', 'd'), [(1, 2, 3, 0), (11, 12, 0, 14)])
可选：存储此数据的另一种方式
您可以将此字典列表重构为列表字典：
def refactor(data):
keys = data[0].keys()
return { k: [d[k] for d in data] for k in keys }
>>> refactor([{"a": 1, "b": 2, "c": 3},{"a": 4, "c": 6, "b": 5}])
{'a': [1, 4], 'b': [2, 5], 'c': [3, 6]}
同样，您可以小心丢失密钥：
def refactor(data):
keys = set(k for d in data for k in d.keys())
return { k: [d[k] for d in data if k in d] for k in keys }
>>> refactor([{'a':1, 'b': 2, 'c': 3}, {'a':11, 'b':12, 'd':14}])
{'d': [14], 'a': [1, 11], 'c': [3], 'b': [2, 12]}
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