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route results from yield to a file

I have the following Python script using Scrapy:
import scrapy
class ChemSpider(scrapy.Spider):
name = "site"
def start_requests(self):
urls = [
'https://www.site.com.au'
]
for url in urls:
yield scrapy.Request(url=url, callback=self.parse)
def parse(self, response):
category_links = response.css('li').xpath('a/#href').getall()
category_links_filtered = [x for x in category_links if 'shop-online' in x] # remove non category links
category_links_filtered = list(dict.fromkeys(category_links_filtered)) # remove duplicates
for category_link in category_links_filtered:
if "medicines" in category_link:
next_page = response.urljoin(category_link) + '?size=10'
self.log(next_page)
yield scrapy.Request(next_page, callback=self.parse_subcategories)
def parse_subcategories(self, response):
for product in response.css('div.Product'):
yield {
'category_link': response.url,
'product_name': product.css('img::attr(alt)').get(),
'product_price': product.css('span.Price::text').get().replace('\n','')
}
My solution will run multiple instances of this script, each scraping a different subset of information from different 'categories'. I know you can run scrapy from the command line to output to a json file, but i want do to the output to a file from within the function, so each instance writes to a different file. Being a beginner with Python, I'm not sure where to go with my script. I need to get the output of the yield into a file while the script is executing. How do i achieve this? There will be hundreds of rows scraped, and I'm not familiar enough with how yield works to understand how to 'return' from it a set of data (or a list) that can then be written to the file.

You are looking to append a file. But being file writing an I/O operation, you need to lock the file from being written by other processes while a process is writing.
Easiest way to achieve is to write in different random files (files with random names) in a directory and concatenating them all using another process.

First let me suggest you some changes to your code. If you want to remove duplicates i you could use a set like this:
category_links_filtered = (x for x in category_links if 'shop-online' in x) # remove non category links
category_links_filtered = set(category_links_filtered) # remove duplicates
note that i'm also changing the [ to ( to make a generator instead of a list and save some memory. Search more about generators: https://www.python-course.eu/python3_generators.php
OK then the solution for your problem is using an Item Pipeline (https://docs.scrapy.org/en/latest/topics/item-pipeline.html), what this does perfom some action on every item yielded from your function parse_subcategories. What you do is add a class in your pipelines.py file and enable this pipeline in settings.py. This is:
In settings.py:
ITEM_PIPELINES = {
'YOURBOTNAME.pipelines.CategoriesPipeline': 300, #the number here is the priority of the pipeline, dont worry and just leave it
}
In pipelines.py:
import json
from urlparse import urlparse #this is library to parse urls
class CategoriesPipeline(object):
#This class dynamically saves the data depending on the category name obtained in the url or by an atrtribute
def open_spider(self, spider):
if hasattr(spider, 'filename'):
#the filename is an attribute set by -a filename=somefilename
filename = spider.filename
else:
#you could also set the name dynamically from the start url like this, if you set -a start_url=https://www.site.com.au/category-name
try:
filename = urlparse(spider.start_url).path[1:] #this returns 'category-name' and replace spaces with _
except AttributeError:
spider.crawler.engine.close_spider(self, reason='no start url') #this should not happen
self.file = open(filename+'.jl', 'w')
def close_spider(self, spider):
self.file.close()
def process_item(self, item, spider):
line = json.dumps(dict(item)) + "\n"
self.file.write(line)
return item
In spiders/YOURBOTNAME.py modify this:
class ChemSpider(scrapy.Spider):
name = "site"
if !hasattr(self, 'start_url'):
spider.crawler.engine.close_spider(self, reason='no start url') #we need a start url
start_urls = [ self.start_url ] #see why this works on https://docs.scrapy.org/en/latest/intro/tutorial.html#a-shortcut-for-creating-requests
def parse(self, response):#...
and then you start your crawl with this command: scrapy crawl site -a start_url=https://www.site.com.au/category-name and you could optionally add -a filename=somename

Python Scrapy not outputting to csv file

What am I doing wrong with the script so it's not outputting a csv file with the data? I am running the script with scrapy runspider yellowpages.py -o items.csv and still nothing is coming out but a blank csv file. I have followed different things here and also watched youtube trying to figure out where I am making the mistake and still cannot figure out what I am not doing correctly.
# -*- coding: utf-8 -*-
import scrapy
import requests
search = "Plumbers"
location = "Hammond, LA"
url = "https://www.yellowpages.com/search"
q = {'search_terms': search, 'geo_location_terms': location}
page = requests.get(url, params=q)
page = page.url
items = ()
class YellowpagesSpider(scrapy.Spider):
name = 'quotes'
allowed_domains = ['yellowpages.com']
start_urls = [page]
def parse(self, response):
self.log("I just visited: " + response.url)
items = response.css('a[class=business-name]::attr(href)')
for item in items:
print(item)

Simple spider without project.
Use my code, I wrote comments to make it easier to understand. This spider looks for all blocks on all pages for a pair of parameters "service" and "location". To run, use:
In your case:
scrapy runspider yellowpages.py -a servise="Plumbers" -a location="Hammond, LA" -o Hammondsplumbers.csv
The code will also work with any queries. For example:
scrapy runspider yellowpages.py -a servise="Doctors" -a location="California, MD" -o MDDoctors.json
etc...
# -*- coding: utf-8 -*-
import scrapy
from scrapy.http import Request
from scrapy.exceptions import CloseSpider
class YellowpagesSpider(scrapy.Spider):
name = 'yellowpages'
allowed_domains = ['yellowpages.com']
start_urls = ['https://www.yellowpages.com/']
# We can use any pair servise + location on our request
def __init__(self, servise=None, location=None):
self.servise = servise
self.location = location
def parse(self, response):
# If "service " and" location " are defined
if self.servise and self.location:
# Create search phrase using "service" and " location"
search_url = 'search?search_terms={}&geo_location_terms={}'.format(self.servise, self.location)
# Send request with url "yellowpages.com" + "search_url", then call parse_result
yield Request(url=response.urljoin(search_url), callback=self.parse_result)
else:
# Else close our spider
# You can add deffault value if you want.
self.logger.warning('=== Please use keys -a servise="service_name" -a location="location" ===')
raise CloseSpider()
def parse_result(self, response):
# all blocks without AD posts
posts = response.xpath('//div[#class="search-results organic"]//div[#class="v-card"]')
for post in posts:
yield {
'title': post.xpath('.//span[#itemprop="name"]/text()').extract_first(),
'url': response.urljoin(post.xpath('.//a[#class="business-name"]/#href').extract_first()),
}
next_page = response.xpath('//a[#class="next ajax-page"]/#href').extract_first()
# If we have next page url
if next_page:
# Send request with url "yellowpages.com" + "next_page", then call parse_result
yield scrapy.Request(url=response.urljoin(next_page), callback=self.parse_result)

for item in items:
print(item)
put yield instead of print there,
for item in items:
yield item

On inspection of your code, I notice a number of problems:
First, you initialize items to a tuple, when it should be a list: items = [].
You should change your name property to reflect the name you want on your crawler so you can use it like so: scrapy crawl my_crawler where name = "my_crawler".
start_urls is supposed to contain strings, not Request objects. You should change the entry from page to the exact search string you want to use. If you have a number of search strings and want to iterate over them, I would suggest using a middleware.
When you try to extract the data from CSS you're forgetting to call extract_all() which would actually transform your selector into string data which you could use.
Also, you shouldn't be redirecting to the standard output stream because a lot of logging goes there and it'll make your output file really messy. Instead, you should extract the responses into items using loaders.
Finally, you're probably missing the appropriate settings from your settings.py file. You can find the relevant documentation here.
FEED_FORMAT = "csv"
FEED_EXPORT_FIELDS = ["Field 1", "Field 2", "Field 3"]

How to pass two user-defined arguments to a scrapy spider

Following How to pass a user defined argument in scrapy spider, I wrote the following simple spider:
import scrapy
class Funda1Spider(scrapy.Spider):
name = "funda1"
allowed_domains = ["funda.nl"]
def __init__(self, place='amsterdam'):
self.start_urls = ["http://www.funda.nl/koop/%s/" % place]
def parse(self, response):
filename = response.url.split("/")[-2] + '.html'
with open(filename, 'wb') as f:
f.write(response.body)
This seems to work; for example, if I run it from the command line with
scrapy crawl funda1 -a place=rotterdam
It generates a rotterdam.html which looks similar to http://www.funda.nl/koop/rotterdam/. I would next like to extend this so that one can specify a subpage, for instance, http://www.funda.nl/koop/rotterdam/p2/. I've tried the following:
import scrapy
class Funda1Spider(scrapy.Spider):
name = "funda1"
allowed_domains = ["funda.nl"]
def __init__(self, place='amsterdam', page=''):
self.start_urls = ["http://www.funda.nl/koop/%s/p%s/" % (place, page)]
def parse(self, response):
filename = response.url.split("/")[-2] + '.html'
with open(filename, 'wb') as f:
f.write(response.body)
However, if I try to run this with
scrapy crawl funda1 -a place=rotterdam page=2
I get the following error:
crawl: error: running 'scrapy crawl' with more than one spider is no longer supported
I don't really understand this error message, as I'm not trying to crawl two spiders, but simply trying to pass two keyword arguments to modify the start_urls. How could I make this work?

When providing multiple arguments you need to prefix -a for every argument.
The correct line for your case would be:
scrapy crawl funda1 -a place=rotterdam -a page=2

Python, Passing data in Scrapy

How do I actually pass data into parse for my spider, le's say variable name or temp.
class CSpider(scrapy.Spider):
name = "s1"
allowed_domains = ["abc.com"]
temp = ""
start_urls = [
url.strip() for url in lists
]
def parse(self, response):
//How do i pass data into here, eg name, temp

If you are defining the temp variable as a class-level variable, you can access it via self.temp.
If this is something you want to be passed from a command-line, see the following topics:
How to give URL to scrapy for crawling?
Scrapy : How to pass list of arguments through command prompt to spider?

As alecxe answered, you can use attributes (class-level variables) to make variables or constants accessible wherever in your class or you can also add a parameter to your method (functions of a class) parse if you want to be able to give values to that parameter that would come from outside of the class.
I'll try here to give you an example of your code with both solutions.
Using an attribute:
class CSpider(scrapy.Spider):
name = "s1"
allowed_domains = ["abc.com"]
temp = ""
# Here is our attribute
self.number_of_days_in_a_week = 7
start_urls = [
url.strip() for url in lists
]
def parse(self, response):
# It is now used in the method
print(f"In a week, there is {self.number_of_days_in_a_week} days.")
If you need to, here is how to pass it as an other argument:
class CSpider(scrapy.Spider):
name = "s1"
allowed_domains = ["abc.com"]
temp = ""
start_urls = [
url.strip() for url in lists
]
def parse(self, what_you_want_to_pass_in):
print(f"In a week, there is {what_you_want_to_pass_in} days.")
# We create an instance of the spider
spider1 = CSpider
# Then we use it's method with an argument
spider1.parse(7)
Note that in the second example, I took back the response argument from your parse method because it was easier to show how the arguments would be passed. Still, if you consider the entire Scrapy framework, you can for sure add external values using this solution.

How can I use the fields_to_export attribute in BaseItemExporter to order my Scrapy CSV data?

I have made a simple Scrapy spider that I use from the command line to export my data into the CSV format, but the order of the data seem random. How can I order the CSV fields in my output?
I use the following command line to get CSV data:
scrapy crawl somwehere -o items.csv -t csv
According to this Scrapy documentation, I should be able to use the fields_to_export attribute of the BaseItemExporter class to control the order. But I am clueless how to use this as I have not found any simple example to follow.
Please Note: This question is very similar to THIS one. However, that question is over 2 years old and doesn't address the many recent changes to Scrapy and neither provides a satisfactory answer, as it requires hacking one or both of:
contrib/exporter/init.py
contrib/feedexport.py
to address some previous issues, that seem to have already been resolved...
Many thanks in advance.

To use such exporter you need to create your own Item pipeline that will process your spider output. Assuming that you have simple case and you want to have all spider output in one file this is pipeline you should use (pipelines.py):
from scrapy import signals
from scrapy.contrib.exporter import CsvItemExporter
class CSVPipeline(object):
def __init__(self):
self.files = {}
#classmethod
def from_crawler(cls, crawler):
pipeline = cls()
crawler.signals.connect(pipeline.spider_opened, signals.spider_opened)
crawler.signals.connect(pipeline.spider_closed, signals.spider_closed)
return pipeline
def spider_opened(self, spider):
file = open('%s_items.csv' % spider.name, 'w+b')
self.files[spider] = file
self.exporter = CsvItemExporter(file)
self.exporter.fields_to_export = [list with Names of fields to export - order is important]
self.exporter.start_exporting()
def spider_closed(self, spider):
self.exporter.finish_exporting()
file = self.files.pop(spider)
file.close()
def process_item(self, item, spider):
self.exporter.export_item(item)
return item
Of course you need to remember to add this pipeline in your configuration file (settings.py):
ITEM_PIPELINES = {'myproject.pipelines.CSVPipeline': 300 }

You can now specify settings in the spider itself.
https://doc.scrapy.org/en/latest/topics/settings.html#settings-per-spider
To set the field order for exported feeds, set FEED_EXPORT_FIELDS.
https://doc.scrapy.org/en/latest/topics/feed-exports.html#feed-export-fields
The spider below dumps all links on a website (written against Scrapy 1.4.0):
import scrapy
from scrapy.http import HtmlResponse
class DumplinksSpider(scrapy.Spider):
name = 'dumplinks'
allowed_domains = ['www.example.com']
start_urls = ['http://www.example.com/']
custom_settings = {
# specifies exported fields and order
'FEED_EXPORT_FIELDS': ["page", "page_ix", "text", "url"],
}
def parse(self, response):
if not isinstance(response, HtmlResponse):
return
a_selectors = response.xpath('//a')
for i, a_selector in enumerate(a_selectors):
text = a_selector.xpath('normalize-space(text())').extract_first()
url = a_selector.xpath('#href').extract_first()
yield {
'page_ix': i + 1,
'page': response.url,
'text': text,
'url': url,
}
yield response.follow(url, callback=self.parse) # see allowed_domains
Run with this command:
scrapy crawl dumplinks --loglevel=INFO -o links.csv
Fields in links.csv are ordered as specified by FEED_EXPORT_FIELDS.

I found a pretty simple way to solve this issue. The above answers I would still say are more correct, but this is a quick fix. It turns out scrapy pulls the items in alphabetical order. Capitals are also important. So, an item beginning with 'A' will be pulled first, then 'B', 'C', etc, followed by 'a', 'b', 'c'. I have a project going right now where the header names are not extremely important, but I did need the UPC to be the first header for input into another program. I have the following item class:
class ItemInfo(scrapy.Item):
item = scrapy.Field()
price = scrapy.Field()
A_UPC = scrapy.Field()
ID = scrapy.Field()
time = scrapy.Field()
My CSV file outputs with the headers (in order): A_UPC, ID, item, price, time