'Scrapy doesn't save all results to csv

I'm trying to scrape a specific web page and although on the console I get all results, on the outputted csv I don't. In this case, I want both title and author of a specific search, but I only get the title. If I reverse the order of the two I get author, so it only takes the first one. Why?

import scrapy

QUERY = "q=brilliant+friend&qt=results_page#x0%253Abook-%2C%2528x0%253Abook%2Bx4%253Aprintbook%2529%2C%2528x0%253Abook%2Bx4%253Adigital%2529%2C%2528x0%253Abook%2Bx4%253Alargeprint%2529%2C%2528x0%253Abook%2Bx4%253Amss%2529%2C%2528x0%253Abook%2Bx4%253Athsis%2529%2C%2528x0%253Abook%2Bx4%253Abraille%2529%2C%2528x0%253Abook%2Bx4%253Amic%2529%2Cx0%253Aartchap-%2C%2528x0%253Aartchap%2Bx4%253Achptr%2529%2C%2528x0%253Aartchap%2Bx4%253Adigital%2529format"

class Spider(scrapy.Spider):
    name = 'worldcatspider'
    start_urls = ['https://www.worldcat.org/search?start=%s&%s' % (number, QUERY) for number in range(0, 4400, 10)]

    def parse(self, response):
        for title in response.css('.name a > strong ::text').extract():
            yield {"title:": title}
        for author in response.css('.author ::text').extract():
            yield {"author:": author}


Solution 1:[1]

My suggestion will be put for statement their head class or div.

I haven't checked but this should work:

def parse(self, response):
 for page in response.css('.menuElem'): 
    title = page.css('.name a > strong ::text').extract()
    author = page.css('.author ::text').extract()
    yield {"title": title,
           "author:": author}

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 Furkan Ozalp