Question

我是Python的新手，我收到了这个错误：

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

在这段代码上：

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

Answer 1

虽然StackOverflow页面中的缩进错误很明显，但它们可能不在您的编辑器中。这里有不同的缩进类型，1,4和8个空格。根据{{3}}，您应该始终使用四个空格进行缩进。你还应该PEP8。

我还建议您尝试使用avoid mixing tabs and spaces运行脚本，以确定何时意外混合制表符和空格。当然，任何体面的编辑器都能够突出显示标签与空格（例如'-tt' command-line option）。

Answer 2

缩进错误，正如错误告诉您的那样。正如您所看到的，您已经缩短了以指示的行开头的代码，而不是for循环中的代码，但是太多而不能与for循环处于同一级别。 Python认为缺少缩进作为结束for循环，然后抱怨你已经过多地缩进了其余的代码。（我所投注的def行只是Stack Overflow希望您格式化代码的工件。）

编辑：鉴于您的更正，我打赌您在源文件中混合使用制表符和空格，这样它就像代码排列的人眼一样，但Python认为它没有。正如其他人所建议的那样，建议的做法是使用空格（参见PEP 8）。如果您使用python -t启动Python，如果代码中有混合标签和空格，您将收到警告，这可以帮助您查明问题。

Answer 3

错误非常简单 - 以check_exists_sql开头的行没有正确缩进。从代码的上下文中，我会缩进它和以下行以匹配它之前的行：

   #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)

继续缩进，直到for循环结束（一直到items.append(item)为止。

Answer 4

由于错误显示您没有正确缩进代码，check_exists_sql未与其上方的行cursor = db.cursor()对齐。

还可以使用4个空格进行缩进。

阅读此http://diveintopython.net/getting_to_know_python/indenting_code.html

Answer 5

import urllib.request
import requests
from bs4 import BeautifulSoup

        r = requests.get('https://icons8.com/icons/set/favicon')

如果您尝试连接到这样的站点，则会出现缩进错误。

import urllib.request
import requests
from bs4 import BeautifulSoup


r = requests.get('https://icons8.com/icons/set/favicon')

Python关心缩进

Answer 6

如果您未正确写入块，则会发生此错误。忘记“：”，或者不使用“制表符”按钮来显示块并使用空格。当您将代码从一个编辑器传输到另一个编辑器时，可能会发生。永远不要忘记这一点：错误并不总是在那一行上。我是来这里的，但是尝试之后我忘记了一个例外。因为我的编辑器不规范，所以发生了。但这在普通编辑器中是可能的。

IndentationError：意外的缩进错误

6 个答案: