[python] IndentationError: unexpected indent error

I am new to Python and am getting this error:

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

On this bit of code:

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?

The indentation is wrong, as the error tells you. As you can see, you have indented the code beginning with the indicated line too little to be in the for loop, but too much to be at the same level as the for loop. Python sees the lack of indentation as ending the for loop, then complains you have indented the rest of the code too much. (The def line I'm betting is just an artifact of how Stack Overflow wants you to format your code.)

Edit: Given your correction, I'm betting you have a mixture of tabs and spaces in the source file, such that it looks to the human eye like the code lines up, but Python considers it not to. As others have suggested, using only spaces is the recommended practice (see PEP 8). If you start Python with python -t, you will get warnings if there are mixed tabs and spaces in your code, which should help you pinpoint the issue.

This error occur when you don't correctly write blocks. Forgetting a ":", or not using "Tab" button for blocks and use spaces. When you are transporting a code from one editor to another editor,it can happen. And never forget this: errors aren't always on that line. I came here for this, but I've forgotten an except after a try. because of my unstandard editor, it happend. But it's possible in normal editor.

import urllib.request
import requests
from bs4 import BeautifulSoup

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

If you try to connect to such a site, you will get an indent error.

import urllib.request
import requests
from bs4 import BeautifulSoup


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

Python cares about indents

As the error says you have not correctly indented code, check_exists_sql is not aligned with line above it cursor = db.cursor() .

Also use 4 spaces for indentation.

Read this http://diveintopython.net/getting_to_know_python/indenting_code.html

The error is pretty straightforward - the line starting with check_exists_sql isn't indented properly. From the context of your code, I'd indent it and the following lines to match the line before it:

   #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)

And keep indenting it until the for loop ends (all the way through to and including items.append(item).