概述
名字：harrY姓氏：PottEr – >用户名：Harry-Potter
名字：HARRY姓氏：POTTER – >用户名：Harry-Potter-1
名字：harrY姓氏：PottEr – >用户名：Harry-Potter-2
等等..
这是我的函数定义：
def return_slug(firstname,lastname):
u_username = firstname.title()+'-'+lastname.title() //Step 1
u_username = '-'.join(u_username.split()) //Step 2
count = User.objects.filter(username=u_username).count() //Step 3
if count==0:
return (u_username)
else:
return (u_username+'-%s' % count)我无法弄清楚在执行第3步之前要做什么.我应该把[：len(u_username)]放在哪里比较字符串？
编辑：
如果存在多个Harry-Potter实例,则通过解决最后添加整数的问题来应用此方法.我的问题是：我将如何检查附加到Harry-Potter的最后一个整数是多少.
from django.utils.text import slugify
def return_slug(firstname,lastname):
# get a slug of the firstname and last name.
# it will normalize the string and add dashes for spaces
# i.e. 'HaRrY POTTer' -> 'harry-potter'
u_username = slugify(unicode('%s %s' % (firstname,lastname)))
# split the username by the dashes,capitalize each part and re-combine
# 'harry-potter' -> 'Harry-Potter'
u_username = '-'.join([x.capitalize() for x in u_username.split('-')])
# count the number of users that start with the username
count = User.objects.filter(username__startswith=u_username).count()
if count == 0:
return u_username
else:
return '%s-%d' % (u_username,count)总结
以上是编程之家为你收集整理的python – Django中的Slugify字符串全部内容，希望文章能够帮你解决python – Django中的Slugify字符串所遇到的程序开发问题。
如果您也喜欢它,动动您的小指点个赞吧
602392714
清零编程群