用文章的标题创建一个网址 [英] Create a url with an article's title
问题描述
localhost：5000 / blog / how-to-use-flask-and-mongodb-无缝地与英雄。 完成这个的最佳方法是什么？任何正确的方向指针赞赏。我不确定从哪里开始。
您正在寻找一种方法来生成slug并使用它来识别帖子。
如果你只想使用一个slu，，所有的帖子标题将必须有一个独特的slu（（这大约意味着一个独特的标题） 。这也意味着，如果你改变帖子的标题，url可能会改变，这将使书签和其他外部链接无效。
更好的方法是做什么像堆栈溢出没有问题。如果你看这个问题的URL，你会注意到它有一个独特的ID和一个slu。。事实上，slug是可选的，你仍然可以通过从url中删除它来到这个页面。
你需要一种方法来生成slu，和一个自定义网址转换器。 变形库提供了一个很好的方式来使用 parameterize 方法。下面的URL转换器需要一个对象，并返回一个URL为 the_object.id 和 the_object.title 作为一个slug。当解析一个url时，它会直接返回对象的 id ，因为slug是可选的。
来自inflection从werkzeug.routing导入参数化
import BaseConverter
$ b $ class IDSlugConverter（BaseConverter）：
匹配一个int id和可选的slug，
$ b：参数attr：slugify的字段名，或者缺省的str（实例）
：参数长度：构建url时的slu max最大长度
regex = r' - ？\ d +（？：/ [\ w\ - ] *）？'
def __init __（self， map，attr ='title'，length = 80）：
self.attr = attr
self.length = int（length）
super（IDSlugConverter，self）.__ init __（map）
$ b $ def to_python（self，value）：
id，slug =（value.split（'/'）+ [None]）[：2]
return int（id）
$ b $ def to_url（self，value）：
raw = str（value）if self.attr is None else getattr（value，self.att （raw）[：self.length] .rstrip（' - '）
return'{} / {}'。format（value.id，slug）。 rstrip（'/'）
注册转换器以便在路由中使用：
$ b $
app.url_map.converters ['id_slug'] = IDSlugConverter
在路线中使用它：
@ app.route（'/ blog / < id_slug：id>'）
def blog_post（id）：
＃通过ID获取帖子，做某事
生成帖子的网址。请注意，您将对象（'post'）（而不仅仅是id）传递给id参数：
url_for blog_post'，id = post）
＃/ blog / 1234 / the-post-title
由我为Stack Overflow编写的转换器Python聊天室 site 。
I have articles in MongoDB. I want the URLs for the articles to be readable. If I have an article named "How to Use Flask and MongoDB Seamlessly with Heroku", I want the URL to be something like localhost:5000/blog/how-to-use-flask-and-mongodb-seamlessly-with-heroku.
What is the best way to accomplish this? Any pointers in the right direction are appreciated. I wasn't sure exactly where to start on this one.
You are looking for a way to generate a "slug" and use that to identify the post.
If you want to use just a slug, all post titles will have to have a unique slug (which approximately means a unique title). This also means that if you change the post's title, the url could change, which would invalidate bookmarks and other outside links.
A better method is to do something like what Stack Overflow does for questions. If you look at this question's URL, you'll notice it has a unique id and a slug. In fact, the slug is optional, you can still get to this page by removing it from the url.
You'll need a way to generate slugs, and a custom url converter. The inflection library provides a nice way to slugify strings with the parameterize method. The following url converter takes an object and returns a url with the_object.id and the_object.title as a slug. When parsing a url, it will just return the object's id, since the slug is optional.
from inflection import parameterize
from werkzeug.routing import BaseConverter
class IDSlugConverter(BaseConverter):
"""Matches an int id and optional slug, separated by "/".
:param attr: name of field to slugify, or None for default of str(instance)
:param length: max length of slug when building url
"""
regex = r'-?\d+(?:/[\w\-]*)?'
def __init__(self, map, attr='title', length=80):
self.attr = attr
self.length = int(length)
super(IDSlugConverter, self).__init__(map)
def to_python(self, value):
id, slug = (value.split('/') + [None])[:2]
return int(id)
def to_url(self, value):
raw = str(value) if self.attr is None else getattr(value, self.attr, '')
slug = parameterize(raw)[:self.length].rstrip('-')
return '{}/{}'.format(value.id, slug).rstrip('/')
Register the converter so it can be used in routes:
app.url_map.converters['id_slug'] = IDSlugConverter
Use it in a route:
@app.route('/blog/<id_slug:id>')
def blog_post(id):
# get post by id, do stuff
Generate a url for a post. Note that you pass the object ('post'), not just the id, to the id parameter.:
url_for('blog_post', id=post)
# /blog/1234/the-post-title
Converter written by me for the Stack Overflow Python chat room site.
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