We Keep Coding

Scrapy spider outputs empy csv file

This is my first question here and I'm learning how to code by myself so please bear with me.
I'm working on a final CS50 project which I'm trying to built a website that aggregates online Spanish course from edx.org and other open online couses websites maybe. I'm using scrapy framework to scrap the filter results of Spanish courses on edx.org... Here is my first scrapy spider which I'm trying to get in each courses link to then get it's name (after I get the code right, also get the description, course url and more stuff).
from scrapy.item import Field, Item
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractor import LinkExtractor
from scrapy.loader import ItemLoader
class Course_item(Item):
name = Field()
#description = Field()
#img_url = Field()
class Course_spider(CrawlSpider):
name = 'CourseSpider'
allowed_domains = ['https://www.edx.org/']
start_urls = ['https://www.edx.org/course/?language=Spanish']
rules = (Rule(LinkExtractor(allow=r'/course'), callback='parse_item', follow='True'),)
def parse_item(self, response):
item = ItemLoader(Course_item, response)
item.add_xpath('name', '//*[#id="course-intro-heading"]/text()')
yield item.load_item()
When I run the spider with "scrapy runspider edxSpider.py -o edx.csv -t csv" I get an empty csv file and I also think is not getting into the right spanish courses results.
Basically I want to get in each courses of this link edx Spanish courses and get the name, description, provider, page url and img url.
Any ideas for why might be the problem?

You can't get edx content with a simple request, it uses javascript rendering for getting the course element dynamically, so CrawlSpider won't work on this case, because you need to find specific elements inside the response body to generate a new Request that will get what you need.
The real request (to get the urls of the courses) is this one, but you need to generate it from the previous response body (although you could just visit it an also get the correct data).
So, to generate the real request, you need data that is inside a script tag:
from scrapy import Spider
import re
import json
class Course_spider(Spider):
name = 'CourseSpider'
allowed_domains = ['edx.org']
start_urls = ['https://www.edx.org/course/?language=Spanish']
def parse(self, response):
script_text = response.xpath('//script[contains(text(), "Drupal.settings")]').extract_first()
parseable_json_data = re.search(r'Drupal.settings, ({.+})', script_text).group(1)
json_data = json.loads(parseable_json_data)
...
Now you have what you need on json_data and only need to create the string URL.

This page use JavaScript to get data from server and add to page.
It uses urls like
https://www.edx.org/api/catalog/v2/courses/course-v1:IDBx+IDB33x+3T2017
Last part is course's number which you can find in HTML
<main id="course-info-page" data-course-id="course-v1:IDBx+IDB33x+3T2017">
Code
from scrapy.http import Request
from scrapy.item import Field, Item
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractor import LinkExtractor
from scrapy.loader import ItemLoader
import json
class Course_spider(CrawlSpider):
name = 'CourseSpider'
allowed_domains = ['www.edx.org']
start_urls = ['https://www.edx.org/course/?language=Spanish']
rules = (Rule(LinkExtractor(allow=r'/course'), callback='parse_item', follow='True'),)
def parse_item(self, response):
print('parse_item url:', response.url)
course_id = response.xpath('//*[#id="course-info-page"]/#data-course-id').extract_first()
if course_id:
url = 'https://www.edx.org/api/catalog/v2/courses/' + course_id
yield Request(url, callback=self.parse_json)
def parse_json(self, response):
print('parse_json url:', response.url)
item = json.loads(response.body)
return item
from scrapy.crawler import CrawlerProcess
c = CrawlerProcess({
'USER_AGENT': 'Mozilla/5.0',
'FEED_FORMAT': 'csv', # csv, json, xml
'FEED_URI': 'output.csv', #
})
c.crawl(Course_spider)
c.start()

from scrapy.http import Request
from scrapy import Spider
import json
class edx_scraper(Spider):
name = "edxScraper"
start_urls = [
'https://www.edx.org/api/v1/catalog/search?selected_facets[]=content_type_exact%3Acourserun&selected_facets[]=language_exact%3ASpanish&page=1&page_size=9&partner=edx&hidden=0&content_type[]=courserun&content_type[]=program&featured_course_ids=course-v1%3AHarvardX+CS50B+Business%2Ccourse-v1%3AMicrosoft+DAT206x+1T2018%2Ccourse-v1%3ALinuxFoundationX+LFS171x+3T2017%2Ccourse-v1%3AHarvardX+HDS2825x+1T2018%2Ccourse-v1%3AMITx+6.00.1x+2T2017_2%2Ccourse-v1%3AWageningenX+NUTR101x+1T2018&featured_programs_uuids=452d5bbb-00a4-4cc9-99d7-d7dd43c2bece%2Cbef7201a-6f97-40ad-ad17-d5ea8be1eec8%2C9b729425-b524-4344-baaa-107abdee62c6%2Cfb8c5b14-f8d2-4ae1-a3ec-c7d4d6363e26%2Ca9cbdeb6-5fc0-44ef-97f7-9ed605a149db%2Cf977e7e8-6376-400f-aec6-84dcdb7e9c73'
]
def parse(self, response):
data = json.loads(response.text)
for course in data['objects']['results']:
url = 'https://www.edx.org/api/catalog/v2/courses/' + course['key']
yield response.follow(url, self.course_parse)
if 'next' in data['objects'] is not None:
yield response.follow(data['objects']['next'], self.parse)
def course_parse(self, response):
course = json.loads(response.text)
yield{
'name': course['title'],
'effort': course['effort'],
}

How to make Scrapy crawl only 1 page (make it non recursive)?

I'm using the latest version of scrapy (http://doc.scrapy.org/en/latest/index.html) and am trying to figure out how to make scrapy crawl only the URL(s) fed to it as part of start_url list. In most cases I want to crawl only 1 page, but in some cases there may be multiple pages that I will specify. I don't want it to crawl to other pages.
I've tried setting the depth level=1 but I'm not sure that in testing it accomplished what I was hoping to achieve.
Any help will be greatly appreciated!
Thank you!
2015-12-22 - Code update:
# -*- coding: utf-8 -*-
import scrapy
from generic.items import GenericItem
class GenericspiderSpider(scrapy.Spider):
name = "genericspider"
def __init__(self, domain, start_url, entity_id):
self.allowed_domains = [domain]
self.start_urls = [start_url]
self.entity_id = entity_id
def parse(self, response):
for href in response.css("a::attr('href')"):
url = response.urljoin(href.extract())
yield scrapy.Request(url, callback=self.parse_dir_contents)
def parse_dir_contents(self, response):
for sel in response.xpath("//body//a"):
item = GenericItem()
item['entity_id'] = self.entity_id
# gets the actual email address
item['emails'] = response.xpath("//a[starts-with(#href, 'mailto')]").re(r'mailto:\s*(.*?)"')
yield item
Below, in the first response, you mention using a generic spider --- isn't that what I'm doing in the code? Also are you suggesting I remove the
callback=self.parse_dir_contents
from the parse function?
Thank you.

looks like you are using CrawlSpider which is a special kind of Spider to crawl multiple categories inside pages.
For only crawling the urls specified inside start_urls just override the parse method, as that is the default callback of the start requests.

Below is a code for the spider that will scrape the title from a blog (Note: the xpath might not be the same for every blog)
Filename: /spiders/my_spider.py
class MySpider(scrapy.Spider):
name = "craig"
allowed_domains = ["www.blogtrepreneur.com"]
start_urls = ["http://www.blogtrepreneur.com/the-best-juice-cleanse-for-weight-loss/"]
def parse(self, response):
hxs = HtmlXPathSelector(response)
dive = response.xpath('//div[#id="tve_editor"]')
items = []
item = DmozItem()
item["title"] = response.xpath('//h1/text()').extract()
item["article"] = response.xpath('//div[#id="tve_editor"]//p//text()').extract()
items.append(item)
return items
The above code will only fetch the title and the article body of the given article.

I got the same problem, because I was using
import scrapy from scrapy.spiders import CrawlSpider
Then I changed to
import scrapy from scrapy.spiders import Spider
And change the class to
class mySpider(Spider):

scrapy crawlspider output

I'm having an issue running through the CrawlSpider example in the Scrapy documentation. It seems to be crawling just fine but I'm having trouble getting it to output to a CSV file (or anything really).
So, my question is can I use this:
scrapy crawl dmoz -o items.csv
or do I have to create an Item Pipeline?
UPDATED, now with code!:
import scrapy
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors import LinkExtractor
from targets.item import TargetsItem
class MySpider(CrawlSpider):
name = 'abc'
allowed_domains = ['ididntuseexample.com']
start_urls = ['http://www.ididntuseexample.com']
rules = (
# Extract links matching 'category.php' (but not matching 'subsection.php')
# and follow links from them (since no callback means follow=True by default).
Rule(LinkExtractor(allow=('ididntuseexample.com', ))),
)
def parse_item(self, response):
self.log('Hi, this is an item page! %s' % response.url)
item = TargetsItem()
item['title'] = response.xpath('//h2/a/text()').extract() #this pulled down data in scrapy shell
item['link'] = response.xpath('//h2/a/#href').extract() #this pulled down data in scrapy shell
return item

Rules are the mechanism CrawlSpider uses for following links. Those links are defined with a LinkExtractor. This element basically indicates which links to extract from the crawled page (like the ones defined in the start_urls list) to be followed. Then you can pass a callback that will be called on each extracted link, or more precise, on the pages downloaded following those links.
Your rule must call the parse_item. So, replace:
Rule(LinkExtractor(allow=('ididntuseexample.com', ))),
with:
Rule(LinkExtractor(allow=('ididntuseexample.com',)), callback='parse_item),
This rule defines that you want to call parse_item on every link whose href is ididntuseexample.com. I suspect that what you want as link extractor is not the domain, but the links you want to follow/scrape.
Here you have a basic example that crawls Hacker News to retrieve the title and the first lines of the first comment for all the news in the main page.
import scrapy
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors import LinkExtractor
class HackerNewsItem(scrapy.Item):
title = scrapy.Field()
comment = scrapy.Field()
class HackerNewsSpider(CrawlSpider):
name = 'hackernews'
allowed_domains = ['news.ycombinator.com']
start_urls = [
'https://news.ycombinator.com/'
]
rules = (
# Follow any item link and call parse_item.
Rule(LinkExtractor(allow=('item.*', )), callback='parse_item'),
)
def parse_item(self, response):
item = HackerNewsItem()
# Get the title
item['title'] = response.xpath('//*[contains(#class, "title")]/a/text()').extract()
# Get the first words of the first comment
item['comment'] = response.xpath('(//*[contains(#class, "comment")])[1]/font/text()').extract()
return item

Scrapy XPath all the links on the page

I am trying to collect all the URLs under a domain using Scrapy. I was trying to use the CrawlSpider to start from the homepage and crawl their web. For each page, I want to use Xpath to extract all the hrefs. And store the data in a format like key-value pair.
Key: the current Url
Value: all the links on this page.
class MySpider(CrawlSpider):
name = 'abc.com'
allowed_domains = ['abc.com']
start_urls = ['http://www.abc.com']
rules = (Rule(SgmlLinkExtractor()), )
def parse_item(self, response):
hxs = HtmlXPathSelector(response)
item = AbcItem()
item['key'] = response.url
item['value'] = hxs.select('//a/#href').extract()
return item
I define my AbcItem() looks like below:
from scrapy.item import Item, Field
class AbcItem(Item):
# key: url
# value: list of links existing in the key url
key = Field()
value = Field()
pass
And when I run my code like this:
nohup scrapy crawl abc.com -o output -t csv &
The robot seems like began to crawl and I can see the nohup.out file being populated by all the configurations log but there is no information from my output file.. which is what I am trying to collect, can anyone help me with this? what might be wrong with my robot?

You should have defined a callback for a rule. Here's an example for getting all links from twitter.com main page (follow=False):
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.item import Item, Field
class MyItem(Item):
url= Field()
class MySpider(CrawlSpider):
name = 'twitter.com'
allowed_domains = ['twitter.com']
start_urls = ['http://www.twitter.com']
rules = (Rule(SgmlLinkExtractor(), callback='parse_url', follow=False), )
def parse_url(self, response):
item = MyItem()
item['url'] = response.url
return item
Then, in the output file, I see:
http://status.twitter.com/
https://twitter.com/
http://support.twitter.com/forums/26810/entries/78525
http://support.twitter.com/articles/14226-how-to-find-your-twitter-short-code-or-long-code
...
Hope that helps.

if you dont set the callback function explicitly, scrapy will use the method parse to process crawled pages. so, you should add parse_item as the callback, or change it's name to parse.

crawling multiple webpages from a website

from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector
class DmozSpider(BaseSpider):
name = "dmoz"
allowed_domains = ["dmoz.org"]
start_urls = [
"www.dmoz.org/Computers/Programming/Languages/Python/Books/",
"www.dmoz.org/Computers/Programming/Languages/Python/Resources/"
]
def parse(self, response):
hxs = HtmlXPathSelector(response)
sites = hxs.select('//ul/li')
for site in sites:
title = site.select('a/text()').extract()
link = site.select('a/#href').extract()
desc = site.select('text()').extract()
print title, link, desc
This is my code. I want plenty of URLs to scrape using loop. So how am I suposed to these? I did put multiple urls in there but I didn't get output from all of them. Some URLs stop responding. So how can I get the data for sure using this code?

You code looks ok but are you sure that start_urls shouldn't start with http://
start_urls = [
"http://www.dmoz.org/Computers/Programming/Languages/Python/Books/",
"http://www.dmoz.org/Computers/Programming/Languages/Python/Resources/"
]
UPD
start_urls is a list of urls scrapy starts with. Usually it has one or two links. Rarely more.
This pages must have identical HTML structure because Scrapy spider process them the same way.
See if i put 4-5 url's in start_urls it gives output ok for first 2-3
url's.
I don't believe this because scrapy doesn't care how many links is start_urls list.
But it stops responding and also tell me how i can implement GUI for this.?
Scrapy has debug shell to test you code.

You just posted the code from the tutorial. What you should do is to actually read the whole documentation, especially the basic concept part. What you basically want is the crawl spider where you can define rules that the spider will follow and process with your given code.
To quote the docs with the example:
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from scrapy.item import Item
class MySpider(CrawlSpider):
name = 'example.com'
allowed_domains = ['example.com']
start_urls = ['http://www.example.com']
rules = (
# Extract links matching 'category.php' (but not matching 'subsection.php')
# and follow links from them (since no callback means follow=True by default).
Rule(SgmlLinkExtractor(allow=('category\.php', ), deny=('subsection\.php', ))),
# Extract links matching 'item.php' and parse them with the spider's method parse_item
Rule(SgmlLinkExtractor(allow=('item\.php', )), callback='parse_item'),
)
def parse_item(self, response):
self.log('Hi, this is an item page! %s' % response.url)
hxs = HtmlXPathSelector(response)
item = Item()
item['id'] = hxs.select('//td[#id="item_id"]/text()').re(r'ID: (\d+)')
item['name'] = hxs.select('//td[#id="item_name"]/text()').extract()
item['description'] = hxs.select('//td[#id="item_description"]/text()').extract()
return item