title: Providing url for spider using scrapyd api
body: <p>I tried something like:</p>
  
  <pre><code>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)
  </code></pre>
  
  <p>And when I check the logs, I got this error code:</p>
  
  <pre><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
  </code></pre>
  
  <p>Looks like only the first letter of "<a href="http://www.foo.com" rel="nofollow">http://www.foo.com</a>" is used as <code>request.url</code>, and I really have no idea why.</p>
  
  <h3>Update</h3>
  
  <p>Maybe start_urls should be a string instead of a list containing 1 element, so I also tried:</p>
  
  <pre><code>"start_urls": "http://www.foo.com"
  </code></pre>
  
  <p>and </p>
  
  <pre><code>"start_urls": [["http://www.foo.com"]]
  </code></pre>
  
  <p>only to get the same error.</p>