如何在Django中创建一个独特的插件 [英] How to create a unique slug in Django

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：
http://www.example.com/buy-a-new-bike_Boston-MA-02111_2

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

我使用这个代码段来生成

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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