[FIXED] How to use scrapy to get "Production Co" from imdb?

Issue

I want to scrap the Production Co from a imdb movie but i dont have any idea how.

I dont know what kinda of information i need to retrive. And right now i only have to get the movie title.

This is my code:
# -*- coding: utf-8 -*-
"""
Created on Sun Jan 27 20:10:53 2019

@author: Razva
"""


import scrapy
from imdb2.items import Imdb2Item


class ThirdSpider(scrapy.Spider):
name = "imdbtestspider"
allowed_domains = ["imdb.com"]
start_urls = (
    'http://www.imdb.com/chart/top',
)

def parse(self, response):
    links = response.xpath('//tbody[@class="lister-list"]/tr/td[@class="titleColumn"]/a/@href').extract()
    i =1 
    for link in links:
        abs_url = response.urljoin(link)
        #
        url_next = '//*[@id="main"]/div/span/div/div/div[2]/table/tbody/tr['+str(i)+']/td[3]/strong/text()'
        rating = response.xpath(url_next).extract()
        if (i <= len(links)):
            i=i+1
        yield scrapy.Request(abs_url, callback = self.parse_indetail, meta={'rating' : rating})



def parse_indetail(self,response):
    item = Imdb2Item()
    #
    item['title'] = response.xpath('//div[@class="title_wrapper"]/h1/text()').extract()[0][:-1]

    return item

If someone can give me a tip, i would aprecciate it.

Solution

For name of Production Co try:

def parse_indetail(self,response):
    item = Imdb2Item()
    item['title'] = response.xpath('//div[@class="title_wrapper"]/h1/text()').extract()[0][:-1]
    item['production'] = response.xpath('//h4[contains(text(), "Production Co")]/following-sibling::a/text()').get()
    return item

Answered By - vezunchik