title: i want to click a link of website using scrapy python
body: <pre><code>import scrapy
  from selenium import webdriver
  
  
  class QuotesSpider(scrapy.Spider):
      name = "quotes"
      start_urls = [
          'http://ozhat-turkiye.com/en/brands/a',
      ]
  </code></pre>
  
  <blockquote>
    <p>I want to click a link </p>
  </blockquote>
  
  <pre><code>    def __init__(self):
          self.drivers = webdriver.Firefox('C:/Program Files (x86)\Mozilla Firefox')
  </code></pre>
  
  <blockquote>
    <p>I want to click a link</p>
  </blockquote>
  
  <pre><code>def parse(self, response):
      for title in response.css('div.tabledivinlineblock a.tablelink50::attr(href)').extract():
          yield {'title': title,
                 'response': response.url
                 }
  
     # i want to click this a tag
      next = self.driver.find_element_by_xpath('//*[@id="maincontent_DataPager"]/a[last()]')
  
      # follow pagination links
      # for href in response.css('span#maincontent_DataPager a:last-child'):
      #
      #     yield response.follow(href, self.parse)
  
      next_page = response.css('span#maincontent_DataPager a:last-child::attr(href)').extract_first().strip()
      if next_page is not None:
          yield response.follow(next_page, callback=self.parse)
  </code></pre>