python - 运行命令 "scrapy crawl quotes"时 scrapy 教程中出现无效语法错误

Question

我正在尝试运行此处给出的代码 https://docs.scrapy.org/en/latest/intro/tutorial.html#our-first-spider但我收到以下错误:

Deeps-MacBook-Pro:tutorial project$ scrapy crawl quotes
2018-07-24 17:16:24 [scrapy.utils.log] INFO: Scrapy 1.5.1 started (bot: tutorial)
2018-07-24 17:16:24 [scrapy.utils.log] INFO: Versions: lxml 4.2.3.0, libxml2 2.9.4, cssselect 1.0.3, parsel 1.5.0, w3lib 1.19.0, Twisted 18.7.0, Python 3.7.0 (default, Jul 23 2018, 20:22:55) - [Clang 9.1.0 (clang-902.0.39.2)], pyOpenSSL 18.0.0 (OpenSSL 1.1.0h  27 Mar 2018), cryptography 2.3, Platform Darwin-17.5.0-x86_64-i386-64bit
2018-07-24 17:16:24 [scrapy.crawler] INFO: Overridden settings: {'BOT_NAME': 'tutorial', 'NEWSPIDER_MODULE': 'tutorial.spiders', 'ROBOTSTXT_OBEY': True, 'SPIDER_MODULES': ['tutorial.spiders']}
Traceback (most recent call last):
  File "/usr/local/bin/scrapy", line 11, in <module>
    sys.exit(execute())
  File "/usr/local/lib/python3.7/site-packages/scrapy/cmdline.py", line 150, in execute
    _run_print_help(parser, _run_command, cmd, args, opts)
  File "/usr/local/lib/python3.7/site-packages/scrapy/cmdline.py", line 90, in _run_print_help
    func(*a, **kw)
  File "/usr/local/lib/python3.7/site-packages/scrapy/cmdline.py", line 157, in _run_command
    cmd.run(args, opts)
  File "/usr/local/lib/python3.7/site-packages/scrapy/commands/crawl.py", line 57, in run
    self.crawler_process.crawl(spname, **opts.spargs)
  File "/usr/local/lib/python3.7/site-packages/scrapy/crawler.py", line 170, in crawl
    crawler = self.create_crawler(crawler_or_spidercls)
  File "/usr/local/lib/python3.7/site-packages/scrapy/crawler.py", line 198, in create_crawler
    return self._create_crawler(crawler_or_spidercls)
  File "/usr/local/lib/python3.7/site-packages/scrapy/crawler.py", line 203, in _create_crawler
    return Crawler(spidercls, self.settings)
  File "/usr/local/lib/python3.7/site-packages/scrapy/crawler.py", line 55, in __init__
    self.extensions = ExtensionManager.from_crawler(self)
  File "/usr/local/lib/python3.7/site-packages/scrapy/middleware.py", line 58, in from_crawler
    return cls.from_settings(crawler.settings, crawler)
  File "/usr/local/lib/python3.7/site-packages/scrapy/middleware.py", line 34, in from_settings
    mwcls = load_object(clspath)
  File "/usr/local/lib/python3.7/site-packages/scrapy/utils/misc.py", line 44, in load_object
    mod = import_module(module)
  File "/usr/local/Cellar/python/3.7.0/Frameworks/Python.framework/Versions/3.7/lib/python3.7/importlib/__init__.py", line 127, in import_module
    return _bootstrap._gcd_import(name[level:], package, level)
  File "<frozen importlib._bootstrap>", line 1006, in _gcd_import
  File "<frozen importlib._bootstrap>", line 983, in _find_and_load
  File "<frozen importlib._bootstrap>", line 967, in _find_and_load_unlocked
  File "<frozen importlib._bootstrap>", line 677, in _load_unlocked
  File "<frozen importlib._bootstrap_external>", line 728, in exec_module
  File "<frozen importlib._bootstrap>", line 219, in _call_with_frames_removed
  File "/usr/local/lib/python3.7/site-packages/scrapy/extensions/telnet.py", line 12, in <module>
    from twisted.conch import manhole, telnet
  File "/usr/local/lib/python3.7/site-packages/twisted/conch/manhole.py", line 154
    def write(self, data, async=False):
                              ^
SyntaxError: invalid syntax

谁能帮我解决这个问题吗？

Answer 1

看起来 scrapy(或您正在使用的版本)与 3.7 不兼容。 “async”成为关键字。