404 link detector with scrapy

0readme.md

List all the broken links on your website

Requirements:

python3 and scrapy (pip install scrapy)

Usage

• scrapy runspider -o items.csv -a site="https://yoursite.org" 1spider.py
• python3 2format_results.py

1spider.py

import scrapy


class BrokenLinksSpider(scrapy.Spider):
    name = 'brokenlink-checker'
    handle_httpstatus_list = [404, 500]


    def __init__(self, site, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self.start_urls = [site]
        self.DOMAIN = site.split('//')[1]


    def parse(self, response):
        if response.status in (404, 500):
            item = {}
            item['url'] = response.url
            item['prev_page'] = response.meta['prev_url']
            item['prev_link_url'] = response.meta['prev_href']
            item['prev_link_text'] = response.meta['prev_link_text']
            item['status'] = response.status

            yield item

        if self.DOMAIN in response.url:
            for link in response.css('a'):
                href = link.xpath('@href').extract()
                text = link.xpath('text()').extract()
                if href: # maybe should show an error if no href
                    yield response.follow(link, self.parse, meta={
                        'prev_link_text': text,
                        'prev_href': href,
                        'prev_url': response.url,
                    })

2format_results.py

import csv, itertools

items = csv.DictReader(open('items.csv'))
for page, links in itertools.groupby(items, lambda item: item['prev_page']):
    if page:
        print('PAGE:', page)
        for line in links:
            print('     LINK TEXT:', line['prev_link_text'])
            print('     LINK URL:', line['prev_link_url'])
            print()
        print()

python3 -m minet.cli telegram channel-messages https://*** > ***.cvs

SyntaxError: invalid syntax