[FIXED] How use XPath to scrape new movies data from website imdb.com?

Issue

I use scrapy for this link. I want crawl information movie from website imdb.com.

When I use code XPath

//td[@class="overview-top"]

I recieve a list information of movies.

Here is my code.

import scrapy
import sys
from imbd.items import ImbdItem
class ImbdSpiderSpider(scrapy.Spider):
    name = "imbd_spider"
    allowed_domains = ["imdb.com"]
    start_urls = ()

    def parse(self, response):

        print response.url
        title_movie= response.xpath('//td[@class="overview-top"]/h4/a/text()').extract()
        length_title = len(title_movie)

        if(length_title == 0):
            return  
        # $x()
        # read block



        #print block[0].xpath('//*[@id="main"]/div/div[2]/div[4]/table/tbody/tr[1]/td[2]/h4/a').extract() 

        #print block[1]
        #print block[1].xpath('//td[@class="overview-top"]/h4/a/text()').extract()
        tree = response.xpath('//td[@class="overview-top"]')
        i = 0
        for block in tree:
            #print table    
            title = block.xpath('//h4[@itemprop="name"]/a/text()').extract()
            # author = block.xpath('//span[@itemprop="director"]/span/a/text()')[i].extract()
            # rate = block.xpath('//div[@class="metascore no_ratings"]/strong/text()')[i].extract()
            # time = block.xpath('//time[@itemprop="duration"]/text()')[i].extract()
            # tag = block.xpath('//span[@itemprop="genre"]/text()').extract()
            # des = block.xpath('//div[@class="outline"]/text()')[i].extract()
            print title
            # print author
            # print rate
            # print time
            # print tag
            # print des
            #i = i + 1

       #    page = response.xpath('//div[@class="sort"]/a')
       #    page2 = page.xpath('//a[text()="Next"]/@href')
       #  nextpage = "http://www.imdb.com" + page2[0].extract() # /movie-coming-soon..

        # yield scrapy.Request(nextpage, self.parse)

    def start_requests(self):
        start = "http://www.imdb.com/movies-coming-soon/2017-12/"
        yield self.make_requests_from_url(start)

Here is my result

I want the terminal to show one title, but this shows all titles in the list.

My code : https://github.com/Takehashi/Scrapy-imbd.com/tree/master

Solution

Just add "." in first to avoid duplicate !

        item['title'] = block.xpath('.//h4[@itemprop="name"]/a/text()').extract()
        item['author'] = block.xpath('.//span[@itemprop="director"]/span/a/text()').extract()
        item['rate'] = block.xpath('.//div[@class="metascore no_ratings"]/strong/text()').extract()
        item['time'] = block.xpath('.//time[@itemprop="duration"]/text()').extract()
        item['tag'] = block.xpath('.//span[@itemprop="genre"]/text()').extract()
        item['des'] = block.xpath('.//div[@class="outline"]/text()').extract()

Answered By - The