I've written a scrapy crawler to fetch me some sweet sweet data and it works. I'm very impressed with myself for the achievement. I even created a jupyter notebook to process the data from the Json file I created.
But I've creatyed the program so that people at work can use it and getting them to navigate to a folder and use command lines isn't going to work so I wanted to make something that I can call on and then process afterwards. But for some reason Scrapy just isnt playing ball. I've found a few bits of help but once the crawl has been completed the json output I've requested doesn't appear. But when i command line it, it shows up.
def parse(self, response):
resp_dict = json.loads(response.body)
f = open(file_name, 'w')
json.dump(resp_dict, f, indent=4)
f.close()
this is the bit that works, sometimes. I just don't understand why it wont give me an output when called from a different script. I've also tried to add this in but I think i'm putting it in the wrong place.
settings = get_project_settings()
settings.set('FEED_FORMAT', 'json')
settings.set('FEED_URI', 'result.json')
I can successfully call the Scrapy Spider, i can see the terminal showing me what's going on. But I just can't get the json output. Literally tearing my hair out now.
process = CrawlerProcess(settings={
"FEEDS": {
"items.json": {"format": "json"},
}
})
process.crawl(HoggleSpider)
process.start()
Related
I am new to Scrapy, and I built a simple spider that scrapes my local news site for titles and amount of comments. It scrapes well, but I have a problem with my language encoding.
I have created a Scrapy project that I then run through anaconda prompt to save the output to a file like so (from the project directory):
scrapy crawl MySpider -o test.csv
When I then open the json file with the following code:
with open('test.csv', 'r', encoding = "L2") as f:
file = f.read()
I also tried saving it to json, opening in excel, changing to different encodings from there ... always unreadable, but the characters differ. I am Czech if that is relevant. I need characters like ěščřžýáíé etc., but it is Latin.
What I get: Varuje pĹ\x99ed
What I want: Varuje před
Here is my spider code. I did not change anything in settings or pipeline, though I tried multiple tips from other threads that do this. I spent 2 hours on this already, browsing stack overflow and documentation and I can't find the solution, it's becoming a headache for me. I'm not a programmer so this may be the reason... anyway:
urls = []
for number in range(1,101):
urls.append('https://www.idnes.cz/zpravy/domaci/'+str(number))
class MySpider(scrapy.Spider):
name = "MySpider"
def start_requests(self):
urls = ['https://www.idnes.cz/zpravy/domaci/']
for url in urls:
yield scrapy.Request(url=url, callback=self.parse_main)
def parse_main(self, response):
articleBlocks = response.xpath('//div[contains(#class,"art")]')
articleLinks = articleBlocks.xpath('.//a[#class="art-link"]/#href')
linksToFollow = articleLinks.extract()
for url in linksToFollow:
yield response.follow(url = url, callback = self.parse_arts)
print(url)
def parse_arts(self, response):
for article in response.css('div#content'):
yield {
'title': article.css('h1::text').get(),
'comments': article.css('li.community-discusion > a > span::text').get(),
}
Scrapy saves feed exports with utf-8 encoding by default.
Opening the file with the correct encoding displays the characters fine.
If you want to change the encoding used, you can do so by using the FEED_EXPORT_ENCODING setting (or using FEEDS instead).
After one more hour of trial and error, I solved this. The problem was not in Scrapy, it was correctly saving in utf-8, the problem was in the command:
scrapy crawl idnes_spider -o test.csv
that I ran to save it. When I run the command:
scrapy crawl idnes_spider -s FEED_URI=test.csv -s FEED_FORMAT=csv
It works.
I'm newbie to the python scrapy.
When I push the 'scrapy crawl name' command, the cmd window does something very busily. But finally, it doesn't spit out any HTML files.
There's seems lots of questions about scrapy not working, but couldn't find one like this case. So I post this question.
This is my codes.
import scrapy
class PostsSpider(scrapy.Spider):
name = "posts"
start_urls = [
'https://blog.scrapinghub.com/page/1/',
'https://blog.scrapinghub.com/page/2/'
]
def parse(self, response):
page = reponse.url.split('/')[-1]
filename = 'posts-%s.html' % page
with open(filename, 'wb') as f:
f.write(response.body)
I went in to 'cd postscrape' where all these files and venv are layed.
And activated the venv by 'call venv\Scripts\activate.bat'.
And finally went 'scrapy crawl posts' on the cmd, in which venv was activated.
As you see, if I go like this, this code should spit out two HTML files 'posts-1.html' and 'posts-2.html'.
Actually the command doesn't return any error message and seems to do somethings busily. But finally, it returns nothing.
What's the problem??
Thank you genius!
There is no need to manually write items to file. You can simply yield items and provide flag -o as follows:
scrapy crawl some_spider -o some_file_name.json
More you can check in the documentation.
You missed one letter 's' in the 'response'.
page = reponse.url.split('/')[-1]
-->
page = response.url.split('/')[-1]
I'm trying to save screenshots of scraped webpages with Scrapy Splash. I've copied and pasted the code found here into my pipeline folder: https://docs.scrapy.org/en/latest/topics/item-pipeline.html
Here's the code from the url:
import scrapy
import hashlib
from urllib.parse import quote
class ScreenshotPipeline(object):
"""Pipeline that uses Splash to render screenshot of
every Scrapy item."""
SPLASH_URL = "http://localhost:8050/render.png?url={}"
async def process_item(self, item, spider):
encoded_item_url = quote(item["url"])
screenshot_url = self.SPLASH_URL.format(encoded_item_url)
request = scrapy.Request(screenshot_url)
response = await spider.crawler.engine.download(request, spider)
if response.status != 200:
# Error happened, return item.
return item
# Save screenshot to file, filename will be hash of url.
url = item["url"]
url_hash = hashlib.md5(url.encode("utf8")).hexdigest()
filename = "{}.png".format(url_hash)
with open(filename, "wb") as f:
f.write(response.body)
# Store filename in item.
item["screenshot_filename"] = filename
return item
I've also followed the instructions for setting up splash found here: https://github.com/scrapy-plugins/scrapy-splash
When I call the command scrapy crawl spidereverything works correctly except the pipeline.
This is the "Error" I'm seeing.
<coroutine object ScreenshotPipeline.process_item at 0x7f29a9c7c8c0>
The spider is yielding the item correctly, but it will not process the item.
Does anyone have any advice? Thank you.
Edit:
I think what is going on is that Scrapy is calling the process_item() method as you normally would. However according to these docs: https://docs.python.org/3/library/asyncio-task.html a coroutine object must be called differently.
asyncio.run(process_item()) rather than process_item().
I think I may have to modify the source code?
You should use scrapy-splash in a script inside spider not in the pipelines.
I followed this docs and it works for me.
Both Scrapy and Django Frameworks are standalone best framework of Python to build crawler and web applications with less code, Though still whenever You want to create a spider you always have to generate new code file and have to write same piece of code(though with some variation.) I was trying to integrate both. But stuck at a place where i need to send the status 200_OK that spider run successfully, and at the same time spider keep running and when it finish off it save data to database.
Though i know the API are already available with scrapyd. But i Wanted to make it more versatile. That lets you create crawler without writing multiple file. I thought The Crawlrunner https://docs.scrapy.org/en/latest/topics/practices.html would help in this,therefor try this thing also
t Easiest way to run scrapy crawler so it doesn't block the script
but it give me error that the builtins.ValueError: signal only works in main thread
Even though I get the response back from the Rest Framework. But Crawler failed to run due to this error does that mean i need to switch to main thread?
I am doing this with a simple piece of code
spider = GeneralSpider(pk)
runner = CrawlerRunner()
d = runner.crawl(GeneralSpider, pk)
d.addBoth(lambda _: reactor.stop())
reactor.run()
I ran scrapy spider in django view, and sharing my code.
settings_file_path = "scraping.settings" # Scrapy Project Setting
os.environ.setdefault('SCRAPY_SETTINGS_MODULE', settings_file_path)
settings = get_project_settings()
runner = CrawlerRunner(settings)
path = "/path/to/sample.py"
path = url.replace('.py', '')
path = url.replace('/', '.')
file_path = ".SampleSpider".format(path)
SampleSpider = locate(file_path)
d = runner.crawl(SampleSpider)
d.addBoth(lambda _: reactor.stop())
reactor.run()
I hope it's helpful.
I'm working on a Scrapy app, where I'm trying to login to a site with a form that uses a captcha (It's not spam). I am using ImagesPipeline to download the captcha, and I am printing it to the screen for the user to solve. So far so good.
My question is how can I restart the spider, to submit the captcha/form information? Right now my spider requests the captcha page, then returns an Item containing the image_url of the captcha. This is then processed/downloaded by the ImagesPipeline, and displayed to the user. I'm unclear how I can resume the spider's progress, and pass the solved captcha and same session to the spider, as I believe the spider has to return the item (e.g. quit) before the ImagesPipeline goes to work.
I've looked through the docs and examples but I haven't found any ones that make it clear how to make this happen.
This is how you might get it to work inside the spider.
self.crawler.engine.pause()
process_my_captcha()
self.crawler.engine.unpause()
Once you get the request, pause the engine, display the image, read the info from the user& resume the crawl by submitting a POST request for login.
I'd be interested to know if the approach works for your case.
I would not create an Item and use the ImagePipeline.
import urllib
import os
import subprocess
...
def start_requests(self):
request = Request("http://webpagewithcaptchalogin.com/", callback=self.fill_login_form)
return [request]
def fill_login_form(self,response):
x = HtmlXPathSelector(response)
img_src = x.select("//img/#src").extract()
#delete the captcha file and use urllib to write it to disk
os.remove("c:\captcha.jpg")
urllib.urlretrieve(img_src[0], "c:\captcha.jpg")
# I use an program here to show the jpg (actually send it somewhere)
captcha = subprocess.check_output(r".\external_utility_solving_captcha.exe")
# OR just get the input from the user from stdin
captcha = raw_input("put captcha in manually>")
# this function performs the request and calls the process_home_page with
# the response (this way you can chain pages from start_requests() to parse()
return [FormRequest.from_response(response,formnumber=0,formdata={'user':'xxx','pass':'xxx','captcha':captcha},callback=self.process_home_page)]
def process_home_page(self, response):
# check if you logged in etc. etc.
...
What I do here is that I import urllib.urlretrieve(url) (to store the image), os.remove(file) (to delete the previous image), and subprocess.checoutput (to call an external command line utility to solve the captcha). The whole Scrapy infrastructure is not used in this "hack", because solving a captcha like this is always a hack.
That whole calling external subprocess thing could have been one nicer, but this works.
On some sites it's not possible to save the captcha image and you have to call the page in a browser and call a screen_capture utility and crop on an exact location to "cut out" the captcha. Now that is screenscraping.