python - 从亚马逊的“搜索”页面中刮取ASIN

我尝试在亚马逊上抓取ASIN号码。请注意，这与产品详细信息无关(例如:https://www.youtube.com/watch?v=qRVRIh3GZgI)，但这是在您搜索关键字时(在本示例中为“trimmer”，请尝试以下操作):
https://www.amazon.com/s?k=trimmer&ref=nb_sb_noss_2)。结果是很多产品，我能够刮所有标题。

不可见的是ASIN(这是一个唯一的Amazon号)。在检查HTML时，我看到了文本(href)中的链接，其中包含ASIN编号。在下面的示例中，ASIN = B01MSHQ5IQ

<a class="a-link-normal a-text-normal" href="/Philips-Norelco-Groomer-MG3750-50/dp/B01MSHQ5IQ/ref=sr_1_3?keywords=trimmer&amp;qid=1554462204&amp;s=gateway&amp;sr=8-3">

以我的问题结尾:如何获取页面上的所有产品标题和ASIN号？ 例如:

Number     Title                       ASIN
 1       Braun, Beardtrimmer          B07JH1LLYR
 2       TNT Pro Series Waist         B00R84J2PK
 ...     ...                          ...

到目前为止，我正在使用scrapy(但也适用于其他Python解决方案)，并且能够抓取Titles。

到目前为止，我的代码:

首先在命令行中运行:

scrapy startproject tutorial

然后，调整Spider中的文件(请参见示例1)和items.py(请参见示例2)。

例子1

class AmazonProductSpider(scrapy.Spider):
  name = "AmazonDeals"
  allowed_domains = ["amazon.com"]

  #Use working product URL below
  start_urls = [
     "https://www.amazon.com/s?k=trimmer&ref=nb_sb_noss_2"

]
 ## scrapy crawl AmazonDeals -o Asin_Titles.json

  def parse(self, response):
      items = AmazonItem()


      Title = response.css('.a-text-normal').css('::text').extract()
      items['title_Products'] = Title
      yield items

按照@glhr的要求，添加 items.py 代码:

例子2

# -*- coding: utf-8 -*-

# Define here the models for your scraped items
#
# See documentation in:
# http://doc.scrapy.org/en/latest/topics/items.html

import scrapy

class AmazonItem(scrapy.Item):
  # define the fields for your item here like:
  title_Products = scrapy.Field()

最佳答案

您可以通过提取href的<a class="a-link-normal a-text-normal" href="...">属性来获得产品的链接:

Link = response.css('.a-text-normal').css('a::attr(href)').extract()

在链接中，您可以使用正则表达式从链接中提取ASIN号:

(?<=dp/)[A-Z0-9]{10}

上面的正则表达式将匹配10个字符(大写字母或数字)，然后加上dp/。在此处查看演示:https://regex101.com/r/mLMv3k/1

这是parse()方法的有效实现:

def parse(self, response):
    Link = response.css('.a-text-normal').css('a::attr(href)').extract()
    Title = response.css('span.a-text-normal').css('::text').extract()

    # for each product, create an AmazonItem, populate the fields and yield the item
    for result in zip(Link,Title):
        item = AmazonItem()
        item['title_Product'] = result[1]
        item['link_Product'] = result[0]
        # extract ASIN from link
        ASIN = re.findall(r"(?<=dp/)[A-Z0-9]{10}",result[0])[0]
        item['ASIN_Product'] = ASIN
        yield item

这需要用新字段扩展AmazonItem:

class AmazonItem(scrapy.Item):
    # define the fields for your item here like:
    title_Product = scrapy.Field()
    link_Product = scrapy.Field()
    ASIN_Product = scrapy.Field()

样本输出:

{'ASIN_Product': 'B01MSHQ5IQ',
 'link_Product': '/Philips-Norelco-Groomer-MG3750-50/dp/B01MSHQ5IQ',
 'title_Product': 'Philips Norelco Multigroom Series 3000, 13 attachments, '
                  'FFP, MG3750'}
{'ASIN_Product': 'B01MSHQ5IQ',
 'link_Product': '/Philips-Norelco-Groomer-MG3750-50/dp/B01MSHQ5IQ',
 'title_Product': 'Philips Norelco Multi Groomer MG7750/49-23 piece, beard, '
                  'body, face, nose, and ear hair trimmer, shaver, and clipper'}

演示:https://repl.it/@glhr/55534679-AmazonSpider

要将输出写入JSON文件，只需在Spider中指定Feed导出设置:

class AmazonProductSpider(scrapy.Spider):
    name = "AmazonDeals"
    allowed_domains = ["amazon.com"]
    start_urls = ["https://www.amazon.com/s?k=trimmer&ref=nb_sb_noss_2"]
    custom_settings = {
            'FEED_URI' : 'Asin_Titles.json',
            'FEED_FORMAT' : 'json'
    }

关于python - 从亚马逊的“搜索”页面中刮取ASIN，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55534679/

asin

python - 从亚马逊的“搜索”页面中刮取ASIN