我是 Python 新手，遇到了这个错误:

Traceback (most recent call last):
  File "/usr/local/bin/scrapy", line 4, in <module>
    execute()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/cmdline.py", line 130, in execute
    _run_print_help(parser, _run_command, cmd, args, opts)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/cmdline.py", line 96, in _run_print_help
    func(*a, **kw)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/cmdline.py", line 136, in _run_command
    cmd.run(args, opts)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/commands/crawl.py", line 42, in run
    q = self.crawler.queue
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/command.py", line 31, in crawler
    self._crawler.configure()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/crawler.py", line 36, in configure
    self.spiders = spman_cls.from_settings(self.settings)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/spidermanager.py", line 33, in from_settings
    return cls(settings.getlist('SPIDER_MODULES'))
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/spidermanager.py", line 23, in __init__
    for module in walk_modules(name):
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scrapy/utils/misc.py", line 65, in walk_modules
    submod = __import__(fullpath, {}, {}, [''])
  File "/my_crawler/empt/empt/spiders/empt_spider.py", line 59
    check_exists_sql = "SELECT * FROM LINKS WHERE link = '%s' LIMIT 1" % item['link']
    ^
IndentationError: unexpected indent

关于这段代码:

def parse_item(self, response):
    hxs = HtmlXPathSelector(response)
    sites = hxs.select('//a[contains(@href, ".mp3")]/@href').extract()
    items = [ ]

    #for site in sites:
        #link = site.select('a/@href').extract()
        #print site
    for site in sites:
        item = EmptItem()
        item['link'] = site #site.select('a/@href').extract()

        #### DB INSERT ATTEMPT ###
        #MySQL Test

        #open db connection
        db = MySQLdb.connect("localhost","root","str0ng","TESTDB")

        #prepare a cursor object using cursor() method
        cursor = db.cursor()

        #see if any links in the DB match the crawled link
        check_exists_sql = "SELECT * FROM LINKS WHERE link = '%s' LIMIT 1" % item['link']

        cursor.execute(check_exists_sql)

        if cursor.rowcount = 0:
            #prepare SQL query to insert a record into the db.
            sql = "INSERT INTO LINKS ( link ) VALUES ( '%s')" % item['link']

            try:
                #execute the sql command
                cursor.execute(sql)
                #commit your changes to the db
                db.commit()
            except:
                #rollback on error
                db.rollback()

                #fetch a single row using fetchone() method.
                #data = cursor.fetchone()

                #print "Database version: %s " % data

            #disconnect from server
            db.close()

            ### end mysql

        items.append(item)
    return items​

最佳答案

虽然缩进错误在 StackOverflow 页面中很明显，但它们可能不在您的编辑器中。您在这里混合了不同的缩进类型，1、4 和 8 个空格。根据 PEP8，您应该始终使用四个空格进行缩进。 .您还应该 avoid mixing tabs and spaces .

我还建议您尝试使用 '-tt' command-line option 运行您的脚本。以确定您何时不小心混合了制表符和空格。当然，任何体面的编辑器都可以突出显示制表符而不是空格(例如 Vim's 'list' option )。

关于python - 缩进错误 : unexpected indent error，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/3920628/