如何用python格式化url中的字符串？

Question

我可以用1个变量格式化链接（link1），但是当我想用2个变量格式化链接（link2）时，它会抛出一个错误（ValueError:索引94处不支持的格式字符“A”（0x41） 链接2应该是：https://widget.s24.com/applications/1e082c83/widgets/422/products?searchTerm=Topper&origin=https%3A%2F%2Fwww.real.de%2Fproduct%2F324093002%2F

有什么解决办法吗

i = 324093002
b = "Topper"
link_product = 'https://www.real.de/product/%s/'
link_keyword = 'https://widget.s24.com/applications/1e082c83/widgets/422/products?searchTerm=*%s*&origin=https%3A%2F%2Fwww.real.de%2Fproduct%2F*%s*%2F'
link1 = link_product %i
link2 = link_keyword % (i, b)

Answer 1

Char%在字符串格式中有特殊的含义，如%s、%i等，并且有%3A{}格式不正确。您必须使用%%通知Python您需要普通字符%-%%3A{}

i = 324093002
b = "Topper"
link_keyword = 'https://widget.s24.com/applications/1e082c83/widgets/422/products?searchTerm=*%s*&origin=https%%3A%%2F%%2Fwww.real.de%%2Fproduct%%2F*%s*%%2F'

link2 = link_keyword % (i, b)
print(link2)

Answer 2

我建议使用f字符串格式（>；Python 3.6）。有关更多信息，请参见here

i = 324093002
b = "Topper"
link_product = f"https://www.real.de/product/{i}/"
link_keyword = f"https://widget.s24.com/applications/1e082c83/widgets/422/products?searchTerm=*{i}*&origin=https%3A%2F%2Fwww.real.de%2Fproduct%2F*{b}*%2F"

Answer 3

您应该使用urllib，以防止将数据与已经URL编码的URL混合：

from urllib.parse import urlencode, urljoin, urlparse

i = 324093002
b = "Topper"
link_product = urljoin('https://www.real.de/product/', str(i))

link_url = 'https://widget.s24.com/applications/1e082c83/widgets/422/products?{}'
link_params = {'searchTerm': i, 'origin': link_product}

link_keyword = link_url.format(urlencode(link_params))
print(link_keyword)

输出：

https://widget.s24.com/applications/1e082c83/widgets/422/products?searchTerm=324093002&origin=https%3A%2F%2Fwww.real.de%2Fproduct%2F324093002

注：
特别是当b变成使用非URL安全字符的术语时，如Top top或iPhone7+，使用常规字符串格式会把事情搞砸