我正在构建一个网络应用程序，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 是连续的或低数字。例如，我的 URL 如下所示:

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

最佳答案

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

混淆它们并使它们不连续

要使 ID 不连续，您可以说，将原始值乘以某个任意值，添加随机“箔条”作为要丢弃的数字 - 在我的示例中使用简单的模数检查:

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):

1 262 1
2 440 2
3 576 3
4 684 4
5 841 5
6 977 6
7 1197 7
8 1326 8
9 1364 9
10 1528 10

编辑:再想一想，谷壳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，数字可能会使整个数字无效)。

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

把它们放在一起

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)

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

关于python - 为数据库对象获取半长唯一 ID(非顺序)键的最佳方法是什么，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/9877524/