从 URL、RE、python 中提取 Amzon ASIN

Question

I have a huge list of urls with links to Amazon products, this urls have an information contained within that I need that is called ASIN number.

I understand that one of the best ways to extract that information is via Regular Expressions , I found a pattern in the urls that could help

1- https://www.amazon.com/adidas-Melange-Performance-T-Shirt-Charcoal/dp/B07P4LVZNL/ref=sr_1_fkmr1_2?dchild=1&keywords=Adidas+M%C3%A8lange+Tech+T-Shirt+A372&qid=1579685244&sr=8-2-fkmr1

2- https://www.amazon.com/adidas-Originals-Solid-Melange-Purple/dp/B07DXPN7TK/ref=sr_1_fkmr2_1?dchild=1&keywords=Adidas+M%C3%A8lange+Tech+T-Shirt+A372&qid=1579685244&sr=8-1-fkmr2

3- https://www.amazon.com/adidas-Game-Mode-Polo-Multi-Sport/gp/B07R23QGH6/ref=sr_1_fkmr2_2?dchild=1&keywords=Adidas+M%C3%A8lange+Tech+T-Shirt+A372&qid=1579685244&sr=8-2-fkmr2

The respective ASIN numbers are:

1- B07P4LVZNL , located between: dp/B07P4LVZNL/ref=sr_1_f

2- B07DXPN7TK , located between: dp/B07DXPN7TK/ref=sr_1_fkmr2_

3- B07R23QGH6 , located between: gp/B07R23QGH6/ref=sr_1_fkmr2_

I tried this code:

asin = re.match("http[s]?://www.amazon.com(\w+)(.*)/(dp|gp/product)/(?P<asin>\w+).*", href, flags=re.IGNORECASE)

href is the variable where I have stored the urls

But well... It doesn't work quite well, this is the type of result I get:

<re.Match object; span=(0, 175), match='https://www.amazon.com/adidas-Originals-Solid-Mel>
<re.Match object; span=(0, 171), match='https://www.amazon.com/adidas-Game-Mode-Polo-Mult>
<re.Match object; span=(0, 167), match='https://www.amazon.com/adidas-Tech-Tee-Black-X-La>

Thank you for your help

Answer 1

I suggest using

/[dg]p/([^/]+)

It matches /dp/ or /gp/ and then captures into Group 1 any one or more characters other than / .

See the regex demo . In Python :

asin = re.search(r'/[dg]p/([^/]+)', href, flags=re.IGNORECASE)
if asin:
  print(asin.group(1))