I am traversing this URL. In this javascript:ctrl.set_pageReload(1) function making an AJAX call which then loads page data. How can I write my Rule(LinkExtractor(). to make a traverse or is there some other way?
What is AJAX? Its just simply a request to a link with GET or POST method.
You can check it in Inspect-Element view.
Click on the button you are talking about, then see where is the AJAX going to?
Also instead of scraping URLs via Rule(LinkExtractor(), remove start_urls and def parse() method and do this,
def start_requests(self):
yield Request(url= URLE_HERE, callback=self.parse_detail_page)
Related
So, I'm trying to scrape a website with infinite scrolling.
I'm following this tutorial on scraping infinite scrolling web pages: https://blog.scrapinghub.com/2016/06/22/scrapy-tips-from-the-pros-june-2016
But the example given looks pretty easy, it's an orderly JSON object with the data you want.
I want to scrape this https://www.bahiablancapropiedades.com/buscar#/terrenos/venta/bahia-blanca/todos-los-barrios/rango-min=50.000,rango-max=350.000
The XHR response for each page is weird, looks like corrupted html code
This is how the Network tab looks
I'm not sure how to navigate the items inside "view". I want the spider to enter each item and crawl some information for every one.
In the past I've succesfully done this with normal pagination and rules guided by xpaths.
https://www.bahiablancapropiedades.com/buscar/resultados/0
This is XHR url.
While scrolling the page it will appear the 8 records per request.
So do one thing get all records XPath. these records divide by 8. it will appear the count of XHR requests.
do below process. your issue will solve. I get the same issue as me. I applied below logic. it will resolve.
pagination_count = xpath of presented number
value = int(pagination_count) / 8
for pagination_value in value:
url = https://www.bahiablancapropiedades.com/buscar/resultados/+[pagination_value]
pass this url to your scrapy funciton.
It is not corrupted HTML, it is escaped to prevent it from breaking the JSON. Some websites will return simple JSON data and others, like this one, will return the actual HTML to be added.
To get the elements you need to get the HTML out of the JSON response and create your own parsel Selector (this is the same as when you use response.css(...)).
You can try the following in scrapy shell to get all the links in one of the "next" pages:
scrapy shell https://www.bahiablancapropiedades.com/buscar/resultados/3
import json
import parsel
json_data = json.loads(response.text)
sel = parsel.Selector(json_data['view']) # view contains the HTML
sel.css('a::attr(href)').getall()
I am trying to scrape flyers from flipp.com/weekly_ads using Scrapy. Before I can scrape the flyers, I need to input my area code, and search for local flyers (on the site, this is done by clicking a button).
I am trying to input a value, and simualate "clicking a button" using Scrapy.
Initially, I thought that I would be able to use a FormRequest.from_response to search for the form, and input my area code as a value. However, the button is written in javascript, meaning that the form cannot be found.
So, I tried to find the HTTP call via Inspect Element > Developer Tools > Network > XHR to see if any of the calls would load the equivalent flipp page with the new, inputted area code (my area code).
Now, I am very new to Scrapy, and HTTP requests/responses, so I am unsure if the link I found is the correct one (as in, the response with the new area code), or not.
This is the request I found:
https://gateflipp.flippback.com/bf/flipp/data?locale=en-us&postal_code=90210&sid=10775773055673477
I used an arbitrary postal code for the request (90210).
I suspect this is the incorrect request, but in the case that I am wrong, and this is correct:
How do I navigate to - flipp.com/weekly_ads/groceries from this request, while maintaining the new area code?
If this is incorrect:
How do I input a value for a javascript button, and get the result using Scrapy?
import scrapy
import requests
import json
class flippSpider(scrapy.Spider):
name = "flippSpider"
postal_code = "M1T2R8"
start_urls = ["https://flipp.com/weekly_ads"]
def parse(self, response): #Input value and simulate button click
return Request() #Find http call to simulate button click with correct field/value parameters
def parse_formrequest(self, response):
yield scrapy.Request("https://flipp.com/weekly_ads/groceries", callback= self.parse_groceries)
def parse_groceries(self, response):
flyers = []
flyer_names = response.css("class.flyer-name").extract()
for flyer_name in flyer_names:
flyer = FlippspiderItem()
flyer["name"] = flyer_name
flyers.append(flyer)
self.log(flyer["name"])
print(flyer_name)
return flyers
I expected to find the actual javascript button request within the XHR links but the one I found seems to be incorrect.
Edit: I do not want to use Selenium, it's slow, and I do not want a browser to pop up during execution of the spider.
I suspect this is the incorrect request, but in the case that I am wrong, and this is correct:
That is the correct URL to get the data powering that website; the things you see on screen when you go to flipp.com/weekly_ads/groceries is just packaging that data in HTML
How do I navigate to - flipp.com/weekly_ads/groceries from this request, while maintaining the new area code?
I am pretty sure you are asking the wrong question. You don't need to -- and in fact navigating to flipp.com/weekly_ads/groceries will 100% not do what you want anyway. You can observe that when you click on "Groceries", the content changes but the browser does not navigate to any new page, nor does it make a new XHR request. Thus, everything that you need is in that JSON. What is happening is they are using the flyers.*.categories that contains "Groceries" to narrow down the 129 flyers that are returned to just those related to Groceries.
As for "maintaining the new area code," it's a similar "wrong question" because every piece of data that is returned by that XHR is scoped to the postal code in question. Thus, you don't need to re-submit anything, and nor would I expect any data that comes back from your postal_code=90210 request to contain 30309 (or whatever) data.
Believe it or not, you're actually in a great place: you don't need to deal with complicated CSS or XPath queries to liberate the data from its HTML prison: they are kind enough to provide you with an API to their data. You just need to deal with unpacking the content from their structure into your own.
I'm trying to get links from a page with LinkExtractor on a page with infinite scroll. Doing this with
rules = (
Rule(LinkExtractor(allow=".*?(\/nl\/agenda\/).*"), callback='parse_item', follow=True),
)
works. However, this gets called without JavaScript, thus the images are not loading within the page (and their url, which I need). When changing the LinkExtractor to;
rules = (
Rule(LinkExtractor(allow=".*?(\/nl\/agenda\/).*"), callback='parse_item', follow=True, process_links='process_links'),
)
with;
def process_links(self, links):
for link in links:
link.url = "http://localhost:8050/render.html?" + urlencode({ 'url' : link.url })
return links
It only goes to the urls it loads when loading up the page (but it needs to get ALL the links which you can get with scrolling). For some reason it also loads some weird localhost URLs like so;
http://localhost:8050/render.html?url=http%3A%2F%2Flocalhost%3A8050%2Fnl%2Fagenda%2xxxxxx
Which I have no clue why it does that.
Is there a way to execute JavaScript when using the LinkExtractor and Splash, so I can scroll and get all the links before the LinkExtractor gets the links? Only executing JavaScript when following up links from the LinkExtractor would also be enough, but I wouldn't know where to begin to do that.
Link extractor works on the current content not the content that render dynamically. And yes, as you say, for that, you are using splash but splash is used to render JavaScript code while virtual scrolling is never handled in splash, virtual scrolling is more like a network call to fetch new data and append it to the existing HTML. so when you scroll, find a call and then hit that call to get the desired data.
So I've written a spider that extracts certain desired links from a webpage and puts the URL, link text, and other information not necessarily contained in the <a> tag itself, into an item for each link.
How should I pass this item onto another spider which scrapes the URL provided in that item?
This question has been asked many times.
Below are some links on this site that answer your question.
Some answer it directly ie passing items to another function but you may realise that you do not need to do it that way, so other methods are linked to show whats possible.
Using multiple spiders at in the project in Scrapy
Scrapy - parse a page to extract items - then follow and store item url contents
Scrapy: Follow link to get additional Item data?
I am using scrapy tool to scrape content from website, i need help from you guys how to scrape the reponse which is dynamically loaded from ajax.
when content loading from ajax at that mean time url not changing it keep remains same but content would be changed so on that event i need to crawl.
thank you,
G.kavirajan
yield FormRequest('http://addons.prestashop.com/en/modules/featureproduct/ajax-homefeatured.php',
formdata={'type':'new','ajax':'1'},
callback=self.your_callback_method)
bellow are the urls that you can easily catch using fiddler or firebug
this is for featured tab http://addons.prestashop.com/en/modules/featureproduct/ajax-homefeatured.php?ajax=1&type=random
this is for new tab http://addons.prestashop.com/en/modules/featureproduct/ajax-homefeatured.php?ajax=1&type=new
you can request on these url directly to get results you required, although website is using POST request to get data for these url, but i tried with parameter GET request is also working properly