Providing url for spider using scrapyd api

Question

I tried something like:

payload = {"project": settings['BOT_NAME'],
             "spider": crawler_name,
             "start_urls": ["http://www.foo.com"]}
response = requests.post("http://192.168.1.41:6800/schedule.json",
                           data=payload)

And when I check the logs, I got this error code:

File "/usr/lib/pymodules/python2.7/scrapy/spider.py", line 53, in make_requests_from_url
    return Request(url, dont_filter=True)
  File "/usr/lib/pymodules/python2.7/scrapy/http/request/__init__.py", line 26, in __init__
    self._set_url(url)
  File "/usr/lib/pymodules/python2.7/scrapy/http/request/__init__.py", line 61, in _set_url
    raise ValueError('Missing scheme in request url: %s' % self._url)
exceptions.ValueError: Missing scheme in request url: h

It seems that only the first letter " http://www.foo.com " is used as request.url, and I really don't know why.

Update

Maybe start_urls should be a string instead of a list containing 1 element, so I also tried:

"start_urls": "http://www.foo.com"

and

"start_urls": [["http://www.foo.com"]]

to get the same error.

Answer 1

You can change your spider to get an argument urland add it to start_urlson init.

class MySpider(Spider):

    start_urls = []

    def __init__(self, *args, **kwargs):
        super(MySpider, self).__init__(*args, **kwargs)
        self.start_urls.append(kwargs.get('url'))

    def parse(self, response):
        # do stuff

Now yours payloadwill be:

payload = {
    "project": settings['BOT_NAME'],
    "spider": crawler_name,
    "url": "http://www.foo.com"
}