Question

我正在构建一个Web应用程序，我希望我的URL方案看起来像这样：

someurl.com/object/FJ1341lj

目前我只使用我的SQL Alchemy对象中的主键，但问题是我不希望Urls是连续的还是低数字。例如，我的网址如下所示：

someurl.com/object/1
someurl.com/object/2

Answer 1

编码整数

您可以对整数使用可逆编码：

def int_str(val, keyspace):
    """ Turn a positive integer into a string. """
    assert val >= 0
    out = ""
    while val > 0:
        val, digit = divmod(val, len(keyspace))
        out += keyspace[digit]
    return out[::-1]

def str_int(val, keyspace):
    """ Turn a string into a positive integer. """
    out = 0
    for c in val:
        out = out * len(keyspace) + keyspace.index(c)
    return out

快速测试代码：

keyspace = "fw59eorpma2nvxb07liqt83_u6kgzs41-ycdjh" # Can be anything you like - this was just shuffled letters and numbers, but...
assert len(set(keyspace)) == len(keyspace) # each character must occur only once

def test(v):
    s = int_str(v, keyspace)
    w = str_int(s, keyspace)
    print "OK? %r -- int_str(%d) = %r; str_int(%r) = %d" % (v == w, v, s, s, w)

test(1064463423090)
test(4319193500)
test(495689346389)
test(2496486533)

输出

OK? True -- int_str(1064463423090) = 'antmgabi'; str_int('antmgabi') = 1064463423090
OK? True -- int_str(4319193500) = 'w7q0hm-'; str_int('w7q0hm-') = 4319193500
OK? True -- int_str(495689346389) = 'ev_dpe_d'; str_int('ev_dpe_d') = 495689346389
OK? True -- int_str(2496486533) = '1q2t4w'; str_int('1q2t4w') = 2496486533

模糊它们并使它们不连续

要使ID不连续，您可以将原始值乘以某个任意值，将随机“chaff”添加为要丢弃的数字 - 在我的示例中使用简单模数检查：< / p>

def chaffify(val, chaff_size = 150, chaff_modulus = 7):
    """ Add chaff to the given positive integer.
    chaff_size defines how large the chaffing value is; the larger it is, the larger (and more unwieldy) the resulting value will be.
    chaff_modulus defines the modulus value for the chaff integer; the larger this is, the less chances there are for the chaff validation in dechaffify() to yield a false "okay".
    """
    chaff = random.randint(0, chaff_size / chaff_modulus) * chaff_modulus
    return val * chaff_size + chaff

def dechaffify(chaffy_val, chaff_size = 150, chaff_modulus = 7):
    """ Dechaffs the given chaffed value. The chaff_size and chaff_modulus parameters must be the same as given to chaffify() for the dechaffification to succeed.
    If the chaff value has been tampered with, then a ValueError will (probably - not necessarily) be raised. """
    val, chaff = divmod(chaffy_val, chaff_size)
    if chaff % chaff_modulus != 0:
        raise ValueError("Invalid chaff in value")
    return val

for x in xrange(1, 11):
    chaffed = chaffify(x)
    print x, chaffed, dechaffify(chaffed)

输出（随机性）：

编辑：第二个想法，糠的随机性可能不是一个好主意，因为你失去了每个混淆ID的规范性 - 这缺乏随机性但仍然有验证（改变一个如果chaff_val足够大，数字可能会使整数无效。

def chaffify2(val, chaff_val = 87953):
    """ Add chaff to the given positive integer. """
    return val * chaff_val

def dechaffify2(chaffy_val, chaff_val = 87953):
    """ Dechaffs the given chaffed value. chaff_val must be the same as given to chaffify2(). If the value does not seem to be correctly chaffed, raises a ValueError. """
    val, chaff = divmod(chaffy_val, chaff_val)
    if chaff != 0:
        raise ValueError("Invalid chaff in value")
    return val

全部放在一起

document_id = random.randint(0, 1000000)
url_fragment = int_str(chaffify(document_id))
print "URL for document %d: http://example.com/%s" % (document_id, url_fragment)
request_id = dechaffify(str_int(url_fragment))
print "Requested: Document %d" % request_id

输出（随机性）

URL for document 831274: http://example.com/w840pi
Requested: Document 831274

Answer 2

可能比你想要的时间长一点。

Python 2.7.1+ (r271:86832, Apr 11 2011, 18:13:53) 
[GCC 4.5.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import uuid
>>> uuid.uuid4()
UUID('ba587488-2a96-4daa-b422-60300eb86155')
>>> str(uuid.uuid4())
'001f8565-6330-44a6-977a-1cca201aedcc'
>>>

如果你使用sqlalchemy，你可以像这样定义一个类型为uuid的id列

from sqlalchemy import types
from sqlalchemy.databases.mysql import MSBinary
from sqlalchemy.schema import Column
import uuid


class UUID(types.TypeDecorator):
    impl = MSBinary
    def __init__(self):
        self.impl.length = 16
        types.TypeDecorator.__init__(self,length=self.impl.length)

    def process_bind_param(self,value,dialect=None):
        if value and isinstance(value,uuid.UUID):
            return value.bytes
        elif value and not isinstance(value,uuid.UUID):
            raise ValueError,'value %s is not a valid uuid.UUID' % value
        else:
            return None

    def process_result_value(self,value,dialect=None):
        if value:
            return uuid.UUID(bytes=value)
        else:
            return None

    def is_mutable(self):
        return False


id_column_name = "id"

def id_column():
    import uuid
    return Column(id_column_name,UUID(),primary_key=True,default=uuid.uuid4)

如果您使用的是Django，Preet的答案可能更合适，因为很多django的东西都依赖于整个主键。

Answer 3

你总是可以获取id的哈希值，然后用基数62（0-9，a-z，A-Z）基数表示结果数。

import string
import hashlib

def enc(val):
    chars = string.digits + string.letters
    num_chars = len(chars)
    r=''
    while val!= 0:
        r+=chars[val % num_chars]
        val/=num_chars
    return r

def fancy_id(i, hash_truncate=12):
    h = hashlib.sha1(str(i))
    return enc(int(h.hexdigest()[:hash_truncate], 16))

fancy_id(1) # 'XYY6dYFg'
fancy_id(2) # '6jxNvE961'

~~类似地存在解码函数~~您必须将此生成的url id存储在对象中。这样你就可以从你的url id映射回到对象。

Answer 4

根据您的要求，最好的选择是使用itertools.combinations，就像这样

>>> urls=itertools.combinations(string.ascii_letters,6)
>>> 'someurl.com/object/'+''.join(x.next())
'someurl.com/object/abcdek'
>>> 'someurl.com/object/'+''.join(x.next())
'someurl.com/object/abcdel'
>>> 'someurl.com/object/'+''.join(x.next())
'someurl.com/object/abcdem'

获取Database对象的半长唯一ID（非顺序）键的最佳方法是什么

4 个答案:

编码整数

模糊它们并使它们不连续

全部放在一起