Question

我对刮scrap很陌生，今天就是我的第一天。我遇到了最基本的代码问题。我不断收到错误消息，说invalid syntax。这是我的代码，下面是我遇到的错误。我在终端中运行此程序，而不是python本身。

# -*- coding: utf-8 -*-
import scrapy


class QuotesSpider(scrapy.Spider):
    name = 'quotes'
    allowed_domains = ['toscrape.com']
    start_urls = ['http://toscrape.com/random']

    def parse(self, response):
        print(response.url)

2018-09-11 18:30:29 [scrapy.utils.log] INFO: Scrapy 1.5.1 started (bot: scrapybot)
2018-09-11 18:30:29 [scrapy.utils.log] INFO: Versions: lxml 4.2.5.0, libxml2 2.9.8, cssselect 1.0.3, parsel 1.5.0, w3lib 1.19.0, Twisted 18.7.0, Python 3.7.0 (v3.7.0:1bf9cc5093, Jun 26 2018, 23:26:24) - [Clang 6.0 (clang-600.0.57)], pyOpenSSL 18.0.0 (OpenSSL 1.1.0i  14 Aug 2018), cryptography 2.3.1, Platform Darwin-17.7.0-x86_64-i386-64bit
2018-09-11 18:30:29 [scrapy.crawler] INFO: Overridden settings: {'SPIDER_LOADER_WARN_ONLY': True}
Traceback (most recent call last):
  File "/Library/Frameworks/Python.framework/Versions/3.7/bin/scrapy", line 11, in <module>
    sys.exit(execute())
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/cmdline.py", line 150, in execute
    _run_print_help(parser, _run_command, cmd, args, opts)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/cmdline.py", line 90, in _run_print_help
    func(*a, **kw)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/cmdline.py", line 157, in _run_command
    cmd.run(args, opts)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/commands/runspider.py", line 88, in run
    self.crawler_process.crawl(spidercls, **opts.spargs)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/crawler.py", line 170, in crawl
    crawler = self.create_crawler(crawler_or_spidercls)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/crawler.py", line 198, in create_crawler
    return self._create_crawler(crawler_or_spidercls)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/crawler.py", line 203, in _create_crawler
    return Crawler(spidercls, self.settings)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/crawler.py", line 55, in __init__
    self.extensions = ExtensionManager.from_crawler(self)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/middleware.py", line 58, in from_crawler
    return cls.from_settings(crawler.settings, crawler)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/middleware.py", line 34, in from_settings
    mwcls = load_object(clspath)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/utils/misc.py", line 44, in load_object
    mod = import_module(module)
  File "/Library/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 "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/scrapy/extensions/telnet.py", line 12, in <module>
    from twisted.conch import manhole, telnet
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/twisted/conch/manhole.py", line 154
    def write(self, data, async=False):
                              ^
SyntaxError: invalid syntax

Answer 1

async是Python> 3.4的关键字。您必须检查python版本，并且库支持您的python版本。

Python Scrapy不断返回“无效语法”

1 个答案: