Scrapyd部署"属性错误:NoneType对象没有属性'module_name'"

Question

我正在尝试使用scrapy将我的爬虫项目部署到scrapyd实例,但调用commend会返回以下错误:

服务器响应(200):{"status":"error","message":"AttributeError:'NoneType'对象没有属性'module_name'"}

这是我的setup.py来构建部署期间提交的python egg:

from setuptools import setup, find_packages

setup(
    name = 'mycrawler',
    version = '0.1',
    packages = find_packages(),
    install_requires = [
        'scrapy',
        'PyMongo',
            'simplejson',
            'queue'
    ]
)

我的scrapy.cfg:

[settings]
default = mycrawler.settings

[deploy:scrapyd_home_vm]
url = http://192.168.1.2:6800/
project = mycrawler

[deploy:scrapyd_local_vm]
url = http://192.168.38.131:6800/
project = mycrawler

我觉得这与鸡蛋的制作方式有关,但我不确定.我知道当你访问一个应该是对象的属性时,python会抛出这样的错误,但无论出于何种原因,它实际上都是null.我也没有"module_name"属性或任何试图在我自己的代码中引用它的东西.从本地scrapy运行爬虫工作得很好,但部署鸡蛋却没有.

Answer 1

一个相当晚的答案,但我遇到了同样的问题,并找到了解决方案.

通过查看scrapyd自己发出的追溯,可以找到我的问题:

Traceback (most recent call last):
  File "/usr/lib/python2.7/runpy.py", line 162, in _run_module_as_main
    "__main__", fname, loader, pkg_name)
  File "/usr/lib/python2.7/runpy.py", line 72, in _run_code
    exec code in run_globals
  File "/home/vagrant/venv/lib/python2.7/site-packages/scrapyd/runner.py", line 39, in <module>
    main()
  File "/home/vagrant/venv/lib/python2.7/site-packages/scrapyd/runner.py", line 34, in main
    with project_environment(project):
  File "/usr/lib/python2.7/contextlib.py", line 17, in __enter__
    return self.gen.next()
  File "/home/vagrant/venv/lib/python2.7/site-packages/scrapyd/runner.py", line 22, in project_environment
    activate_egg(eggpath)
  File "/home/vagrant/venv/local/lib/python2.7/site-packages/scrapyd/eggutils.py", line 13, in activate_egg
    settings_module = d.get_entry_info('scrapy', 'settings').module_name

AttributeError: 'NoneType' object has no attribute 'module_name'

正如您所看到的,它正在尝试加载scrapy项目的settings_module; 该module_name属性不存在,因为d.get_entry_info返回None对象.

解决方案是检查您setup.py用于生成鸡蛋的那个,并检查对setup的调用是否包含以下行:

packages=find_packages(),
entry_points={'scrapy': ['settings = scraper.settings']},

这里scraper.settings是scrapy项目设置文件的Python模块路径.将其更改为反映您的项目布局的那个,你应该是花花公子..

..如果没有,这里的关键是检查输出scrapyd(启用调试允许您在服务器响应中看到)以找到解决方案.