在Django中创建一个独特的

在Django中创建一个独特的

本文介绍了如何在Django中创建一个独特的插件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我试图在Django中创建一个独特的s o,以便我可以通过这样的URL访问一个帖子:



相关模型:

 类ZipCode(models.Model):
zipcode = models.CharField max_length = 5)
city = models.CharField(max_length = 64)
statecode = models.CharField(max_length = 32)

类需要(models.Model):
title = models.CharField(max_length = 50)
us_zip = models.CharField(max_length = 5)
slug =

def get_city():
zip = ZipCode.objects.get(zipcode = self.us_zip)
city =%s,%s%s%(zip。城市,zip.statecode,zip.zipcode)
返回城市

示例ZipCode记录:




  • zipcode =02111

  • city =Boston

  • statecode =MA



一个示例需求记录:




  • title =买新自行车

  • us_zip =02111

  • slug =buy-a -new-bike_Boston-MA-02111_2(需要)



有关如何创建这个独特的s>的提示?它的组成是:




  • Need.title +_+ Need.get_city()+_+一个可选的递增整数它独特所有的空格都应该被替换为 - 。



注意:自行车-Boston-MA-02111已经存在,这是它附加了_2,使其独一无二。



我尝试过django扩展,但是似乎只能采取领域或元组的领域来构建独特的s。。我需要传递get_city()函数以及标题和城市之间的_连接器。任何人解决了这个并愿意分享?



谢谢!



更新 / p>

我已经在其UUIDField中使用django扩展,所以如果它也可以用于其AutoSlugField,那将是很好的!

解决方案

我使用这个来生成



slug将是Django SlugField,其中blank = True,但在save方法中执行slug。



需要模型的典型保存方法可能在下面

  def save(self,** kwargs):
slug_str =%s%s%(self.title,self.us_zip)
unique_slugify(self,slug_str)
超级(需要,自).save(** kwargs)

,这样会产生一个sㄧlike like lug lug bike on on on on on--------- -bike_Boston-MA-02111-1等。输出可能有所不同,但您可以随时浏览片段并自定义您的需求。


I am trying to create a unique slug in Django so that I can access a post via a url like this:http://www.example.com/buy-a-new-bike_Boston-MA-02111_2

The relevant models:

class ZipCode(models.Model):
    zipcode = models.CharField(max_length=5)
    city = models.CharField(max_length=64)
    statecode = models.CharField(max_length=32)

class Need(models.Model):
    title = models.CharField(max_length=50)
    us_zip = models.CharField(max_length=5)
    slug = ?????

    def get_city():
        zip = ZipCode.objects.get(zipcode=self.us_zip)
        city = "%s, %s %s" % (zip.city, zip.statecode, zip.zipcode)
        return city

A sample ZipCode record:

  • zipcode = "02111"
  • city = "Boston"
  • statecode = "MA"

A sample Need record:

  • title = "buy a new bike"
  • us_zip = "02111"
  • slug = "buy-a-new-bike_Boston-MA-02111_2" (desired)

Any tips as to how to create this unique slug? Its composition is:

  • Need.title + "_" + Need.get_city() + "_" + an optional incrementing integer to make it unique. All spaces should be replaced with "-".

NOTE: My desired slug above assumes that the slug "buy-a-new-bike_Boston-MA-02111" already exists, which is what it has the "_2" appended to it to make it unique.

I've tried django-extensions, but it seems that it can only take a field or tuple of fields to construct the unique slug. I need to pass in the get_city() function as well as the "_" connector between the title and city. Anyone solved this and willing to share?

Thank you!

UPDATE

I'm already using django-extensions for its UUIDField, so it would be nice if it could also be usable for its AutoSlugField!

解决方案

I use this snippet for generating unique slug and my typical save method look like below

slug will be Django SlugField with blank=True but enforce slug in save method.

typical save method for Need model might look below

def save(self, **kwargs):
    slug_str = "%s %s" % (self.title, self.us_zip)
    unique_slugify(self, slug_str)
    super(Need, self).save(**kwargs)

and this will generate slug like buy-a-new-bike_Boston-MA-02111 , buy-a-new-bike_Boston-MA-02111-1 and so on. Output might be little different but you can always go through snippet and customize to your needs.

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07-30 23:12