How to give URL to scrapy for crawling?

Question

I want to use scrapy for crawling web pages. Is there a way to pass the start URL from the terminal itself?

It is given in the documentation that either the name of the spider or the URL can be given, but when i given the url it throws an error:

//name of my spider is example, but i am giving url instead of my spider name(It works fine if i give spider name).

scrapy crawl example.com

ERROR:

File "/usr/local/lib/python2.7/dist-packages/Scrapy-0.14.1-py2.7.egg/scrapy/spidermanager.py", line 43, in create raise KeyError("Spider not found: %s" % spider_name) KeyError: 'Spider not found: example.com'

How can i make scrapy to use my spider on the url given in the terminal??

yes example.com is added to allowed_domains. What i really want is to give start_url from command line. how can i do it? — G Gill, Mar 13 '12 at 10:38

score 58 · Accepted Answer · answered Mar 13 '12 at 11:00

58

I'm not really sure about the commandline option. However, you could write your spider like this.

class MySpider(BaseSpider):

    name = 'my_spider'    

    def __init__(self, *args, **kwargs): 
      super(MySpider, self).__init__(*args, **kwargs) 

      self.start_urls = [kwargs.get('start_url')]

And start it like: scrapy crawl my_spider -a start_url="http://some_url"

answered Mar 13 '12 at 11:00

Sjaak Trekhaak

4,596
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thank you so much, this is exactly what i was looking for. It worked fine for me :) – G Gill Mar 13 '12 at 11:26
This approach only works for exactly one url. If you want to provide more than one url, see [my approach](http://stackoverflow.com/a/12749782/1125413) in this thread. – pemistahl Oct 05 '12 at 16:11
2

For multiple URLs: `self.start_urls = kwargs.pop('start_urls').split(',')` which is run *before* the super(). – Steven Almeroth Feb 16 '15 at 18:20

score 17 · Answer 2 · answered Mar 08 '13 at 10:34

An even easier way to allow multiple url-arguments than what Peter suggested is by giving them as a string with the urls separated by a comma, like this:

-a start_urls="http://example1.com,http://example2.com"

In the spider you would then simply split the string on ',' and get an array of urls:

self.start_urls = kwargs.get('start_urls').split(',')

score 8 · Answer 3 · answered Mar 15 '12 at 11:49

8

Use scrapy parse command. You can parse a url with your spider. url is passed from command.

$ scrapy parse http://www.example.com/ --spider=spider-name

http://doc.scrapy.org/en/latest/topics/commands.html#parse

answered Mar 15 '12 at 11:49

Subhash

313
1
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Unfortunately, scrapy parse doesn't seem to have options to save results to a file (in various formats) like scrapy crawl does – dan3 Feb 24 '13 at 07:28
If you are looking to just debug why a particular url your spider is failing on this is an easy option. – jeffjv Apr 20 '16 at 00:10
Can't save/export to file easily. Otherwise this would have been perfect. – Citricguy May 22 '17 at 12:34

score 7 · Answer 4 · answered Feb 16 '15 at 18:20

Sjaak Trekhaak has the right idea and here is how to allow multiples:

class MySpider(scrapy.Spider):
    """
    This spider will try to crawl whatever is passed in `start_urls` which
    should be a comma-separated string of fully qualified URIs.

    Example: start_urls=http://localhost,http://example.com
    """
    def __init__(self, name=None, **kwargs):
        if 'start_urls' in kwargs:
            self.start_urls = kwargs.pop('start_urls').split(',')
        super(Spider, self).__init__(name, **kwargs)

the question arises which is better practice: `kwargs.pop('start_urls')` or `kwargs.get('start_urls')`? — C.W., Nov 08 '20 at 13:30

score 4 · Answer 5 · edited May 23 '17 at 12:16

This is an extension to the approach given by Sjaak Trekhaak in this thread. The approach as it is so far only works if you provide exactly one url. For example, if you want to provide more than one url like this, for instance:

-a start_url=http://url1.com,http://url2.com

then Scrapy (I'm using the current stable version 0.14.4) will terminate with the following exception:

error: running 'scrapy crawl' with more than one spider is no longer supported

However, you can circumvent this problem by choosing a different variable for each start url, together with an argument that holds the number of passed urls. Something like this:

-a start_url1=http://url1.com 
-a start_url2=http://url2.com 
-a urls_num=2

You can then do the following in your spider:

class MySpider(BaseSpider):

    name = 'my_spider'    

    def __init__(self, *args, **kwargs): 
        super(MySpider, self).__init__(*args, **kwargs) 

        urls_num = int(kwargs.get('urls_num'))

        start_urls = []
        for i in xrange(1, urls_num):
            start_urls.append(kwargs.get('start_url{0}'.format(i)))

        self.start_urls = start_urls

This is a somewhat ugly hack but it works. Of course, it's tedious to explicitly write down all command line arguments for each url. Therefore, it makes sense to wrap the scrapy crawl command in a Python subprocess and generate the command line arguments in a loop or something.

Hope it helps. :)

If I call scrapy 0.24.4 like this: `scrapy crawl MySpider -a start_urls=http://example.com/ -o - -t json` Everything work well. Initially I put options between -o and - and get same as You error. — mmv-ru, Nov 02 '15 at 22:11

score 0 · Answer 6 · answered Aug 28 '15 at 12:20

0

You can also try this:

>>> scrapy view http://www.sitename.com

It will open a window in browser of requested URL.

answered Aug 28 '15 at 12:20

Mayur Koshti

1,582
13
20

How to give URL to scrapy for crawling?

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