我有一个脚本以下列方式获取DNS(CNAME,MX,NS)数据:
from dns import resolver
...
def resolve_dns(url):
response_dict = {}
print "\nResolving DNS for %s" % (url)
try:
response_dict['CNAME'] = [rdata for rdata in resolver.query(url, 'CNAME')]
except:
pass
try:
response_dict['MX'] = [rdata for rdata in resolver.query(url, 'MX')]
except:
pass
try:
response_dict['NS'] = [rdata for rdata in resolver.query(url, 'NS')]
except:
pass
return response_dict
Run Code Online (Sandbox Code Playgroud)
对于连续的URL,顺序调用此函数.如果可能的话,我想通过同时获取多个URL的数据来加快上述过程.
有没有办法完成上面的脚本为一批URL做的事情(可能返回一个dict对象列表,每个dict对应一个特定URL的数据)?
您可以将工作放入线程池中.你resolve_dns连续3次请求,所以我创建了一个稍微更通用的工作者,它只执行1个查询并用于collections.product生成所有组合.在线程池中,我将chunksize设置为1以减少线程池批处理,如果某些查询需要很长时间,这会增加执行时间.
import dns
from dns import resolver
import itertools
import collections
import multiprocessing.pool
def worker(arg):
"""query dns for (hostname, qname) and return (qname, [rdata,...])"""
try:
url, qname = arg
rdatalist = [rdata for rdata in resolver.query(url, qname)]
return qname, rdatalist
except dns.exception.DNSException, e:
return qname, []
def resolve_dns(url_list):
"""Given a list of hosts, return dict that maps qname to
returned rdata records.
"""
response_dict = collections.defaultdict(list)
# create pool for querys but cap max number of threads
pool = multiprocessing.pool.ThreadPool(processes=min(len(url_list)*3, 60))
# run for all combinations of hosts and qnames
for qname, rdatalist in pool.imap(
worker,
itertools.product(url_list, ('CNAME', 'MX', 'NS')),
chunksize=1):
response_dict[qname].extend(rdatalist)
pool.close()
return response_dict
url_list = ['example.com', 'stackoverflow.com']
result = resolve_dns(url_list)
for qname, rdatalist in result.items():
print qname
for rdata in rdatalist:
print ' ', rdata
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
1954 次 |
| 最近记录: |