scrapy set the output file in code - python

I am using scrapy with python
I can set the output json file in the cmd. but now I need to do that in code.
I tried this:
in the setting
FEED_EXPORTERS = {
'jsonlines': 'scrapy.contrib.exporter.JsonLinesItemExporter',
}
FEED_FORMAT = 'jsonlines'
in the spider
def __init(self):
settings.overrides['FEED_URI'] = 'output.json'
Note
I am developing a simple spider, so I just need Item Exporter, I don't need to create any item pipeline.
Thanks for helping

The answer is found in an example on the Scrapy documentation. You can output to any format by writing the correct item pipeline, as follows:
import json
class JsonWriterPipeline(object):
def __init__(self):
self.file = open('items.jl', 'wb')
def process_item(self, item, spider):
line = json.dumps(dict(item)) + "\n"
self.file.write(line)
return item
Note that you must also include this pipeline in the default Scrapy project settings file.

Related

How to run multiple spiders through individual pipelines?

Total noob just getting started with scrapy.
In my directory structure I have like this...
#FYI: running on Scrapy 2.4.1
WebScraper/
Webscraper/
spiders/
spider.py # (NOTE: contains spider1 and spider2 classes.)
items.py
middlewares.py
pipelines.py # (NOTE: contains spider1Pipeline and spider2Pipeline)
settings.py # (NOTE: I wrote here:
#ITEM_PIPELINES = {
# 'WebScraper.pipelines.spider1_pipelines': 300,
# 'WebScraper.pipelines.spider2_pipelines': 300,
#}
scrapy.cfg
And spider2.py resembles...
class OneSpider(scrapy.Spider):
name = "spider1"
def start_requests(self):
urls = ["url1.com",]
yield scrapy.Request(
url="http://url1.com",
callback=self.parse
)
def parse(self,response):
## Scrape stuff, put it in a dict
yield dictOfScrapedStuff
class TwoSpider(scrapy.Spider):
name = "spider2"
def start_requests(self):
urls = ["url2.com",]
yield scrapy.Request(
url="http://url2.com",
callback=self.parse
)
def parse(self,response):
## Scrape stuff, put it in a dict
yield dictOfScrapedStuff
With pipelines.py looking like...
class spider1_pipelines(object):
def __init__(self):
self.csvwriter = csv.writer(open('spider1.csv', 'w', newline=''))
self.csvwriter.writerow(['header1', 'header2'])
def process_item(self, item, spider):
row = []
row.append(item['header1'])
row.append(item['header2'])
self.csvwrite.writerow(row)
class spider2_pipelines(object):
def __init__(self):
self.csvwriter = csv.writer(open('spider2.csv', 'w', newline=''))
self.csvwriter.writerow(['header_a', 'header_b'])
def process_item(self, item, spider):
row = []
row.append(item['header_a']) #NOTE: this is not the same as header1
row.append(item['header_b']) #NOTE: this is not the same as header2
self.csvwrite.writerow(row)
I have a question about running spider1 and spider2 on different urls with one terminal command:
nohup scrapy crawl spider1 -o spider1_output.csv --logfile spider1.log & scrapy crawl spider2 -o spider2_output.csv --logfile spider2.log
Note: this is an extension of a previous question specific to this stack overflow post (2018).
Desired result: spider1.csv with data from spider1, spider2.csv with data from spider2.
Current result: spider1.csv with data from spider1, spider2.csv BREAKS but error log contains spider2 data, and that there was a keyerror ['header1'], even though the item for spider2 does not include header1, it only includes header_a.
Does anyone know how to run one spider after the other on different urls, and plug data fetched by spider1, spider2, etc. into pipelines specific to that spider, as in spider1 -> spider1Pipeline -> spider1.csv, spider2 -> spider2Pipelines -> spider2.csv.
Or perhaps this is a matter of specifying the spider1_item and spider2_item from items.py? I wonder if I can specify where to insert spider2's data that way.
Thank you!
You can implement this using custom_settings spider attribute to set settings individually per spider
#spider2.py
class OneSpider(scrapy.Spider):
name = "spider1"
custom_settings = {
'ITEM_PIPELINES': {'WebScraper.pipelines.spider1_pipelines': 300}
...
class TwoSpider(scrapy.Spider):
name = "spider2"
custom_settings = {
'ITEM_PIPELINES': {'WebScraper.pipelines.spider2_pipelines': 300}
...

how can I access a variable parameter at spider class from pipelines.py

I have 3 number of spider files and classes. And I want to save item informations at csv file which has different filename defendant the variable parameter of searching condition. For that, I need to access the spider class parameter.
So, my questions are three.
How can I access the spider class's parameter?
What is the best way to make each csv files? The trigger condition is that will call request at parse function for new searching result.
logger = logging.getLogger(__name__) it's not working in pipelines.py
How can I print that information?
Bellow is my log code style
logger.log(logging.INFO,'\n======= %s ========\n', filename)
I had been searching the ways in google so many times. But I couldn't find the solution.
I did try to use from_crawler function, but I couldn't find the adapt case
Scrapy 1.6.0
python 3.7.3
os window 7 / 32bit
Code:
class CensusGetitemSpider(scrapy.Spider):
name = 'census_getitem'
startmonth=1
filename = None
def parse(self, response):
for x in data:
self.filename = str(startmonth+1)
.
.
.
yield item
yield scrapy.Request(link, callback=self.parse)
you can access spider class and instance attributes from pipeline.py using the spider parameter passed in most of pipeline methods.
For example, :
open_spider(self, spider):
self.filename = spider.name
You can see more about item pipelines here https://docs.scrapy.org/en/latest/topics/item-pipeline.html
You can save it directly from the command line, just define a filename:
scrapy crawl yourspider -o output.csv
But if you really need it to be set from the spider, you can use a custom setting per spider, for example:
class YourSpider(scrapy.Spider):
name = 'yourspider'
start_urls = 'www.yoursite.com'
custom_settings = {
'FEED_URI':'output.csv',
'FEED_FORMAT': 'csv',
}
Use spider.logger.info('Your message')

Order a json by field using scrapy

I have created a spider to scrape problems from projecteuler.net. Here I have concluded my answer to a related question with
I launch this with the command scrapy crawl euler -o euler.json and it outputs an array of unordered json objects, everyone corrisponding to a single problem: this is fine for me because I'm going to process it with javascript, even if I think resolving the ordering problem via scrapy can be very simple.
But unfortunately, ordering items to write in json by scrapy (I need ascending order by id field) seem not to be so simple. I've studied every single component (middlewares, pipelines, exporters, signals, etc...) but no one seems useful for this purpose. I'm arrived at the conclusion that a solution to solve this problem doesn't exist at all in scrapy (except, maybe, a very elaborated trick), and you are forced to order things in a second phase. Do you agree, or do you have some idea? I copy here the code of my scraper.
Spider:
# -*- coding: utf-8 -*-
import scrapy
from eulerscraper.items import Problem
from scrapy.loader import ItemLoader
class EulerSpider(scrapy.Spider):
name = 'euler'
allowed_domains = ['projecteuler.net']
start_urls = ["https://projecteuler.net/archives"]
def parse(self, response):
numpag = response.css("div.pagination a[href]::text").extract()
maxpag = int(numpag[len(numpag) - 1])
for href in response.css("table#problems_table a::attr(href)").extract():
next_page = "https://projecteuler.net/" + href
yield response.follow(next_page, self.parse_problems)
for i in range(2, maxpag + 1):
next_page = "https://projecteuler.net/archives;page=" + str(i)
yield response.follow(next_page, self.parse_next)
return [scrapy.Request("https://projecteuler.net/archives", self.parse)]
def parse_next(self, response):
for href in response.css("table#problems_table a::attr(href)").extract():
next_page = "https://projecteuler.net/" + href
yield response.follow(next_page, self.parse_problems)
def parse_problems(self, response):
l = ItemLoader(item=Problem(), response=response)
l.add_css("title", "h2")
l.add_css("id", "#problem_info")
l.add_css("content", ".problem_content")
yield l.load_item()
Item:
import re
import scrapy
from scrapy.loader.processors import MapCompose, Compose
from w3lib.html import remove_tags
def extract_first_number(text):
i = re.search('\d+', text)
return int(text[i.start():i.end()])
def array_to_value(element):
return element[0]
class Problem(scrapy.Item):
id = scrapy.Field(
input_processor=MapCompose(remove_tags, extract_first_number),
output_processor=Compose(array_to_value)
)
title = scrapy.Field(input_processor=MapCompose(remove_tags))
content = scrapy.Field()
If I needed my output file to be sorted (I will assume you have a valid reason to want this), I'd probably write a custom exporter.
This is how Scrapy's built-in JsonItemExporter is implemented.
With a few simple changes, you can modify it to add the items to a list in export_item(), and then sort the items and write out the file in finish_exporting().
Since you're only scraping a few hundred items, the downsides of storing a list of them and not writing to a file until the crawl is done shouldn't be a problem to you.
By now I've found a working solution using pipeline:
import json
class JsonWriterPipeline(object):
def open_spider(self, spider):
self.list_items = []
self.file = open('euler.json', 'w')
def close_spider(self, spider):
ordered_list = [None for i in range(len(self.list_items))]
self.file.write("[\n")
for i in self.list_items:
ordered_list[int(i['id']-1)] = json.dumps(dict(i))
for i in ordered_list:
self.file.write(str(i)+",\n")
self.file.write("]\n")
self.file.close()
def process_item(self, item, spider):
self.list_items.append(item)
return item
Though it may be non optimal, because the guide suggests in another example:
The purpose of JsonWriterPipeline is just to introduce how to write item pipelines. If you really want to store all scraped items into a JSON file you should use the Feed exports.

Needed a little twitch in my scrapy code to shake off redundant data

I wrote a code in scrapy to scrape coffee shops from yellowpage. The total data is around 870 but I'm getting around 1200 with a minimum number of duplicates. Moreover, in the csv output the data are getting placed in every alternate row. Expecting someone to provide me with a workaround. Thanks in advance.
Folder Name "yellpg" and "items.py" contains
from scrapy.item import Item, Field
class YellpgItem(Item):
name= Field()
address = Field()
phone= Field()
Spider Name "yellsp.py" which contains:
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor
from yellpg.items import YellpgItem
class YellspSpider(CrawlSpider):
name = "yellsp"
allowed_domains = ["yellowpages.com"]
start_urls = (
'https://www.yellowpages.com/search?search_terms=Coffee%20Shops&geo_location_terms=Los%20Angeles%2C%20CA&page=1',
)
rules = (Rule(LinkExtractor(allow=('\&page=.*',)),callback='parse_item',follow=True),)
def parse_item(self, response):
page=response.xpath('//div[#class="info"]')
for titles in page:
item = YellpgItem()
item["name"] = titles.xpath('.//span[#itemprop="name"]/text()').extract()
item["address"] = titles.xpath('.//span[#itemprop="streetAddress" and #class="street-address"]/text()').extract()
item["phone"] = titles.xpath('.//div[#itemprop="telephone" and #class="phones phone primary"]/text()').extract()
yield item
To get the CSV output, the command line I'm using:
scrapy crawl yellsp -o items.csv
I could recommend creating a pipeline that stores items to later check if the new ones are duplicates, but that isn't a real solution here, as it could create memory problems.
The real solution here is that you should avoid "storing" duplicates in your final database.
Define what field of your item is going to behave as the index in your database and everything should be working.
The best way would be to use CSVItemExporter in your pipeline. Create a file named pipeline.py inside your scrapy project and add below code lines.
from scrapy import signals
from scrapy.exporters import CsvItemExporter
class CSVExportPipeline(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_coffer_shops.csv' % spider.name, 'w+b') # hard coded filename, not a good idea
self.files[spider] = file
self.exporter = CsvItemExporter(file)
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
Now add these lines in setting.py
ITEM_PIPELINES = {
'your_project_name.pipelines.CSVExportPipeline': 300
}
This custom CSVItemExporter will export your data in CSV styles. If you are not getting the data as expected you can modify process_item method to suits your need.

Scrapy - Output to Multiple JSON files

I am pretty new to Scrapy. I am looking into using it to crawl an entire website for links, in which I would output the items into multiple JSON files. So I could then upload them to Amazon Cloud Search for indexing. Is it possible to split the items into multiple files instead of having just one giant file in the end? From what I've read, the Item Exporters can only output to one file per spider. But I am only using one CrawlSpider for this task. It would be nice if I could set a limit to the number of items included in each file, like 500 or 1000.
Here is the code I have set up so far (based off the Dmoz.org used in the tutorial):
dmoz_spider.py
import scrapy
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor
from tutorial.items import DmozItem
class DmozSpider(CrawlSpider):
name = "dmoz"
allowed_domains = ["dmoz.org"]
start_urls = [
"http://www.dmoz.org/",
]
rules = [Rule(LinkExtractor(), callback='parse_item', follow=True)]
def parse_item(self, response):
for sel in response.xpath('//ul/li'):
item = DmozItem()
item['title'] = sel.xpath('a/text()').extract()
item['link'] = sel.xpath('a/#href').extract()
item['desc'] = sel.xpath('text()').extract()
yield item
items.py
import scrapy
class DmozItem(scrapy.Item):
title = scrapy.Field()
link = scrapy.Field()
desc = scrapy.Field()
Thanks for the help.
I don't think built-in feed exporters support writing into multiple files.
One option would be to export into a single file in jsonlines format basically, one JSON object per line which is convenient to pipe and split.
Then, separately, after the crawling is done, you can read the file in the desired chunks and write into separate JSON files.
So I could then upload them to Amazon Cloud Search for indexing.
Note that there is a direct Amazon S3 exporter (not sure it helps, just FYI).
You can add a name to each item and use a custom pipeline to output to different json files. like so:
from scrapy.exporters import JsonItemExporter
from scrapy import signals
class MultiOutputExporter(object):
#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):
self.items = ['item1','item2']
self.files = {}
self.exporters = {}
for item in self.items:
self.files[item] = open(f'{item}.json', 'w+b')
self.exporters[item] = JsonItemExporter(self.files[item])
self.exporters[item].start_exporting()
def spider_closed(self, spider):
for item in self.items:
self.exporters[item].finish_exporting()
self.files[item].close()
def process_item(self, item, spider):
self.exporters[item.name].export_item()
return item
Then add names to your items as follows:
class Item(scrapy.Item):
name = 'item1'
Now enable the pipeline in scrapy.setting and voila.

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