'scrapy spider code not running because of syntax?

so my projects seem to keep failing for the same reason. I get syntax error. I'm using anaconda and visual code studio, I have the environment setup correctly, i think*.

The code i'm using is the following:

import scrapy
from scrapy.linkextractors import LinkExtractor
from scrapy.spiders import CrawlSpider, Rule


class BestMoviesSpider(CrawlSpider):
    name = 'best_movies'
    allowed_domains = ['imdb.com']
    start_urls = ['https://www.imdb.com/chart/top']

    rules = (
        Rule(LinkExtractor(restrict_xpaths="//td[@class='titleColumn']/a"), callback='parse_item', follow=True),
    )

    def parse_item(self, response):
       yield {
           'title': response.xpath("//h1/text()").get(),      
            'year': response.xpath("//li[@class="ipc-inline-list__item"]/span/text()").get(),    
           'duration': response.xpath("(//li[@class="ipc-inline-list__item"])[3]/text()").get(),     
           'genre': response.xpath("//span[@class="ipc-chip__text"]/text()").get(),     
           'rating': response.xpath("//span[@class="AggregateRatingButton__RatingScore-sc-1ll29m0-1 iTLWoV"]/text()").get(),     
           'movie_url': response.url,     
       }

The error I'm getting is : line 18 'year': response.xpath("//li[@class="ipc-inline-list__item"]/span/text()").get(), ^ SyntaxError: invalid syntax

Also, I have 2 errors on VSC regarding { and ( not being closed but I think that's because my code isn't running.

Thank you in advance!

Solution 1:^[1]

The issue is in the string definition of the xpaths.

You can just use single quotes and you should be fine:

# not
'year': response.xpath("//li[@class="ipc-inline-list__item"]/span/text()").get(),

# Instead use
'year': response.xpath('//li[@class="ipc-inline-list__item"]/span/text()').get(),

Solution 2:^[2]

It seems to be a problem of quotes:

Try to replace year: response.xpath("//li[@class="ipc-inline-list__item"]/span/text()").get()

by year: response.xpath('//li[@class="ipc-inline-list__item"]/span/text()').get()

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	Davide Laghi
Solution 2	Takamura

'scrapy spider code not running because of syntax?

Solution 1:[1]

Solution 2:[2]

Sources

Related Questions

Solution 1:^[1]

Solution 2:^[2]