在 Python/Django 中生成唯一字符串
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Generate a Unique String in Python/Django
提问by Yax
What I want is to generate a string(key) of size 5 for my users on my website. More like a BBM PIN.
我想要的是在我的网站上为我的用户生成一个大小为 5 的字符串（密钥）。更像是 BBM PIN。
The key will contain numbers and uppercase English letters:
密钥将包含数字和大写英文字母：
- AU1B7
- Y56AX
- M0K7A
- AU1B7
- Y56AX
- M0K7A
How can I also be at rest about the uniqueness of the strings even if I generate them in millions?
即使我生成数以百万计的字符串，我怎么能对字符串的唯一性保持冷静？
In the most pythonic way possible, how can I do this?
以最 Pythonic 的方式，我该怎么做？
采纳答案by zzart
My favourite is
我最喜欢的是
import uuid
uuid.uuid4().hex[:6].upper()
If you using django you can set the unique constrain on this field in order to make sure it is unique. https://docs.djangoproject.com/en/dev/ref/models/fields/#django.db.models.Field.unique
如果您使用 django，您可以在此字段上设置唯一约束以确保它是唯一的。https://docs.djangoproject.com/en/dev/ref/models/fields/#django.db.models.Field.unique
回答by nu11p01n73R
Am not sure about any short cryptic ways, but it can be implemented using a simple straight forward function assuming that you save all the generated strings in a set:
我不确定任何简短的神秘方式，但它可以使用一个简单的直接函数来实现，假设您将所有生成的字符串保存在一个集合中：
import random
def generate(unique):
chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890"
while True:
value = "".join(random.choice(chars) for _ in range(5))
if value not in unique:
unique.add(value)
break
unique = set()
for _ in range(10):
generate(unique)
回答by Tianqi Tong
size = 5
''.join(random.choice(string.letters[26:] + string.digits) for in range(size))
this will generate some short code, but they can be duplicated. so check if they are unique in your database before saving.
这将生成一些短代码，但它们可以被复制。所以在保存之前检查它们在您的数据库中是否唯一。
def generate(size=5):
code = ''.join(random.choice(string.letters[26:] + string.digits) for in range(size))
if check_if_duplicate(code):
return generate(size=5)
return code
or using django unique constrain, and handle exceptions.
或使用 django 唯一约束，并处理异常。
回答by Michael Stachura
From 3.6 You can use secrets module to generate nice random strings. https://docs.python.org/3/library/secrets.html#module-secrets
从 3.6 开始，您可以使用 secrets 模块生成漂亮的随机字符串。 https://docs.python.org/3/library/secrets.html#module-secrets
import secrets
print(secrets.token_hex(5))
回答by Caveman
If you can afford to lose '8' and '9' in the generated numbers there is a very pythonic solution to getting a truly random number.
如果您可以承受在生成的数字中丢失 '8' 和 '9'，那么有一个非常pythonic 的解决方案来获得一个真正的随机数。
import os
import base64
base64.b32encode(os.urandom(3))[:5].decode('utf-8')
Since you are going for uniqueness then you have a problem since 36 * 36 * 36 * 36 * 36 = 60'466'176which will definitely result in collisions if you have millions. Since sets are faster than dicts we do...
由于您要追求唯一性，因此您会遇到问题，因为36 * 36 * 36 * 36 * 36 = 60'466'176如果您有数百万，这肯定会导致冲突。由于集合比字典快，我们这样做......
some_dict = set()
def generate():
return base64.b32encode(os.urandom(3))[:5].decode('utf-8')
def generate_unique():
string = generate()
while string not in some_set:
string = generate()
some_set.add(string)
return string
回答by vedant
There is a function in django that does what you're looking for (credits to this answer):
django 中有一个函数可以满足您的要求（归功于此答案）：
Django provides the function
get_random_string()which will satisfy the alphanumeric string generation requirement. You don't need any extra package because it's in thedjango.utils.cryptomodule.>>> from django.utils.crypto import get_random_string >>> unique_id = get_random_string(length=32) >>> unique_id u'rRXVe68NO7m3mHoBS488KdHaqQPD6Ofv'You can also vary the set of characters with
allowed_chars:>>> short_genome = get_random_string(length=32, allowed_chars='ACTG') >>> short_genome u'CCCAAAAGTACGTCCGGCATTTGTCCACCCCT'
Django 提供了
get_random_string()满足字母数字字符串生成要求的函数。您不需要任何额外的包，因为它在django.utils.crypto模块中。>>> from django.utils.crypto import get_random_string >>> unique_id = get_random_string(length=32) >>> unique_id u'rRXVe68NO7m3mHoBS488KdHaqQPD6Ofv'您还可以使用以下命令更改字符集
allowed_chars：>>> short_genome = get_random_string(length=32, allowed_chars='ACTG') >>> short_genome u'CCCAAAAGTACGTCCGGCATTTGTCCACCCCT'
回答by mevaka
I have a unique field, named 'systemCode' within a lot of my models. And I am generating this manually, but also sometimes it can take value from user input, so I have to check this value before saving and if it matches , regenerating this value as a unique value.
我的许多模型中有一个独特的字段，名为“ systemCode”。而且我是手动生成的，但有时它也可以从用户输入中获取值，所以我必须在保存之前检查这个值，如果匹配，则将此值重新生成为唯一值。
And this is how I generate unique strings at this scenario :
这就是我在这种情况下生成唯一字符串的方式：
This is my standard class Model :
这是我的标准类 Model ：
class ClassOne(models.Model):
name = models.CharField(max_length=100)
systemCode = models.CharField(max_length=25, blank=True, null=True, unique=True)
....
I am using save()method to generate and check this systemCodeis unique :
我正在使用save()方法生成并检查此systemCode是否唯一：
def save(self, *args, **kwargs):
systemCode = self.systemCode
if not systemCode:
systemCode = uuid.uuid4().hex[:6].upper()
while ClassOne.objects.filter(systemCode=systemCode).exclude(pk=self.pk).exists():
systemCode = uuid.uuid4().hex[:6].upper()
self.systemCode = systemCode
super(ClassOne, self).save(*args, **kwargs)
But I have same systemCodefield in all my Models. So I am using a function to generate value.
但是我的所有模型中都有相同的systemCode字段。所以我正在使用一个函数来产生价值。
So, this is how to generate unique value for all models using saveSystemCode()function :
因此，这是使用saveSystemCode()函数为所有模型生成唯一值的方法：
import uuid
def saveSystemCode(inClass, inCode, inPK, prefix):
systemCode = inCode
if not systemCode:
systemCode = uuid.uuid4().hex[:6].upper()
while inClass.objects.filter(systemCode=systemCode).exclude(pk=inPK).exists():
systemCode = uuid.uuid4().hex[:6].upper()
return systemCode
class ClassOne(models.Model):
name = models.CharField(max_length=100)
systemCode = models.CharField(max_length=25, blank=True, null=True, unique=True)
....
def save(self, *args, **kwargs):
self.systemCode = saveSystemCode(ClassOne, self.systemCode, self.pk, 'one_')
super(ClassOne, self).save(*args, **kwargs)
class ClassTwo(models.Model):
name = models.CharField(max_length=100)
systemCode = models.CharField(max_length=25, blank=True, null=True, unique=True)
....
def save(self, *args, **kwargs):
self.systemCode = saveSystemCode(ClassTwo, self.systemCode, self.pk, 'two_')
super(ClassTwo, self).save(*args, **kwargs)
class ClassThree(models.Model):
name = models.CharField(max_length=100)
systemCode = models.CharField(max_length=25, blank=True, null=True, unique=True)
....
def save(self, *args, **kwargs):
self.systemCode = saveSystemCode(ClassThree, self.systemCode, self.pk, 'three_')
super(ClassThree, self).save(*args, **kwargs)
whileloop in the 'saveSystemCode' function is preventing to save same value again.
' saveSystemCode' 函数中的while循环阻止再次保存相同的值。
回答by Reza Abbasi
A more secure and shorter way of doing is using Django's crypto module.
一种更安全、更短的方法是使用 Django 的加密模块。
from django.utils.crypto import get_random_string
code = get_random_string(5)
