CODEWITHSUNDEEP
pythonpython-3.xscrapyweb-crawler
Lior Magen
Lior Magen

Reputation: 1573

how to extract asin from an amazon product page

I have the following webpage Product page and I'm trying to get the ASIN from it (in this case ASIN=B014MHZ90M) and I don't have a clue on how to get it from the page.

I'm using Python 3.4, Scrapy and the following code:

hxs = Selector(response)
product_name = "".join(hxs.xpath('//span[contains(@class,"a-text-ellipsis")]/a/text()').extract())
product_model = hxs.xpath('//body//div[@id="buybox_feature_div"]//form[@method="post"]/input[@id="ASIN"/text()').extract()

In this way I don't get the required field (the ASIN number).

  1. What should I do in order to get the product model (ASIN)?

2.Is there a way to debug such code (I'm using PyCharm). I could not use debugger but only run it without seeing what's going on there in 'slow motion'.

Upvotes: 3

Views: 5352

Answers (5)

JustAGuy
JustAGuy

Reputation: 5941

https://www.amazon.com/gp/seller/asin-upc-isbn-info.html

Amazon Standard Identification Numbers (ASINs) are unique blocks of 10 letters and/or numbers that identify items.

Your best option and probably the easiest one is to run a regex on the URL looking for a 10 char string between two "/".

'/\w{10}/'

You can then simply omit the "/"s from the result.

Upvotes: 0

Ignacio Pérez
Ignacio Pérez

Reputation: 672

I use this:

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

Upvotes: 2

user4093955
user4093955

Reputation:

You can get that from the url.

r = re.search('www.amazon.com/dp/(.+)/', response.url)
print r.group(1)

Upvotes: 0

Steve
Steve

Reputation: 976

Looking at the Amazon page you linked, the ASIN number appears in the "Product Details" section. Using the scrapy shell the following xpath

response.xpath('//li[contains(.,"ASIN: ")]//text()').extract()

returns

[u'ASIN: ', u'B014MHZ90M']

For debugging XPATHs I always use scrapy shell and Firebug for Firefox.

Upvotes: 3

user4125604
user4125604

Reputation:

you can extract B014MHZ90M from the response.url

response.url.split("/dp/")[1]

response.url.split("/dp/")[1] = B014MHZ90M

response.url.split("/dp/")[0] = http://www.amazon.com

Upvotes: 4

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