Python 包含请求的URL的爬网列表。get
我正在尝试对CSV文件中包含的URL列表进行爬网。URL列在CSV的第6列中。URL的格式为: 我没有用下面的代码正确读取CSV中的数据。我在哪里犯了编码错误Python 包含请求的URL的爬网列表。get,python,csv,Python,Csv,我正在尝试对CSV文件中包含的URL列表进行爬网。URL列在CSV的第6列中。URL的格式为: 我没有用下面的代码正确读取CSV中的数据。我在哪里犯了编码错误 list_of_urls = open(filename).read() for i in range(6,len(list_of_urls)): try: url=str(list_of_urls[i][0]) #crawl urls secondCrawlRequest =
list_of_urls = open(filename).read()
for i in range(6,len(list_of_urls)):
try:
url=str(list_of_urls[i][0])
#crawl urls
secondCrawlRequest = requests.get(url, headers=http_headers, timeout=5)
raw_html = secondCrawlRequest.text
except requests.ConnectionError as e:
logging.exception(e)
except requests.HTTPError as e:
logging.exception(e)
except requests.Timeout as e:
logging.exception(e)
except requests.RequestException as e:
logging.exception(e)
sys.exit(1)
你应使用:
如果需要跳过标题行,可以通过调用next(reader)
:
你应使用:
如果需要跳过标题行,可以通过调用next(reader)
:
如果url对于csv中的列或行没有固定的引用,您可以简单地使用正则表达式并按如下方式逐行读取文件:
import re
import requests
filename = 'shitty_url.csv'
with open(filename, 'r') as csvfile:
for line in csvfile:
url_pattern = re.search('https:\/\/(.+?) ', line)
if url_pattern:
found_url = url_pattern.group(1)
url = 'https://%s' % found_url
crawler = requests.get(url, timeout=5)
希望这有帮助:)如果url在csv中没有固定的列或行出现,您可以简单地使用regex并按如下方式逐行读取文件:
import re
import requests
filename = 'shitty_url.csv'
with open(filename, 'r') as csvfile:
for line in csvfile:
url_pattern = re.search('https:\/\/(.+?) ', line)
if url_pattern:
found_url = url_pattern.group(1)
url = 'https://%s' % found_url
crawler = requests.get(url, timeout=5)
希望这有帮助:)URL从CSV中的第二行开始。如何绕过标题行?@lifeicomplex添加URL从CSV中的第二行开始。如何绕过标题行?@lifeicomplex添加
import re
import requests
filename = 'shitty_url.csv'
with open(filename, 'r') as csvfile:
for line in csvfile:
url_pattern = re.search('https:\/\/(.+?) ', line)
if url_pattern:
found_url = url_pattern.group(1)
url = 'https://%s' % found_url
crawler = requests.get(url, timeout=5)