【发布时间】:2010-08-02 23:29:44
【问题描述】:
这是我的代码:
from xgoogle.search import GoogleSearch, SearchError
import urllib, urllib2, sys, argparse
global stringArr
stringArr = ["string 1",
"string 2",
"string 3",
"string etc"]
def searchIt(url):
try:
if(args.verbose>='1'): print "[INFO] Opening URL: "+url
response = urllib.urlopen(url)
except urllib2.URLError, e:
print "[ERROR] "+e.reason
return False
except KeyboardInterrupt:
print "Suspended by user..."
sys.exit()
if(checkForStr(response.read())):
if(args.verbose=='0'): print "[INFO] String found in URL: "+url
else:
if(args.verbose>='1'): print "[INFO] No string found in URL: "+url
def checkForStr(html):
global stringArr
try:
if any(checkStr in html for checkStr in stringArr):
return True
else:
return False
except KeyboardInterrupt:
print "Suspended by user..."
sys.exit()
def main():
try:
i=0
gs = GoogleSearch(args.keyword)
gs.results_per_page = 100
results = []
while True:
tmp = gs.get_results()
i = i+1 # page number
if not tmp: # no more results (pages) were found
break
results.extend(tmp)
for r in results: # process results for page
searchIt(r.url) # check for string
del results[:] # clean results
# finished
except SearchError, e:
print "[ERROR] Search failed: %s" % e
except KeyboardInterrupt:
print "Suspended by user..."
sys.exit()
if __name__ == '__main__':
try:
parser = argparse.ArgumentParser()
parser.add_argument('-v', dest='verbose', default='0', help='Verbosity level', choices='012')
group = parser.add_argument_group('required arguments')
group.add_argument('-k', dest='keyword', help='Keyword to use on google query', required=True)
args = parser.parse_args()
main()
except KeyboardInterrupt:
print "Suspended by user..."
sys.exit()
为了便于阅读,我将它缩短了一点,但它应该仍然可以使用。此代码将成为更大脚本的一部分。
我正在使用这个库:XGOOGLE 从 google 抓取结果,然后我访问每个结果以搜索网站是否包含 stringArr 中的任何字符串。
我第一次测试没有任何问题(我在不到 10 个结果后按 ctrl+C),但是第一次让它运行时,在测试了大约 100 个 url 后我得到了这个错误:
File "./StringScan.py", line 99, in <module>
main()
File "./StringScan.py", line 83, in main
checkForStr(r.url)
File "./StringScan.py", line 39, in checkForStr
response = urllib.urlopen(url)
File "/usr/lib/python2.6/urllib.py", line 86, in urlopen
return opener.open(url)
File "/usr/lib/python2.6/urllib.py", line 205, in open
return getattr(self, name)(url)
File "/usr/lib/python2.6/urllib.py", line 344, in open_http
h.endheaders()
File "/usr/lib/python2.6/httplib.py", line 904, in endheaders
self._send_output()
File "/usr/lib/python2.6/httplib.py", line 776, in _send_output
self.send(msg)
File "/usr/lib/python2.6/httplib.py", line 735, in send
self.connect()
File "/usr/lib/python2.6/httplib.py", line 716, in connect
self.timeout)
File "/usr/lib/python2.6/socket.py", line 500, in create_connection
for res in getaddrinfo(host, port, 0, SOCK_STREAM):
IOError: [Errno socket error] [Errno -2] Name or service not known
(行号不一样,因为我修改了代码贴在这里)
之后,我拿回了我的 linux 终端,就像脚本已经完成一样。但我注意到我的电脑运行得不太好,我检查了系统监视器,发现进程 Python 使用了 1.3gb 内存,我不得不终止进程才能让我的电脑恢复正常。
是我的代码中的某些东西导致了这种情况还是为什么会发生?
我知道我的代码可能有一些错误,但现在我主要对可能导致内存问题的任何错误感兴趣。任何帮助将不胜感激。
【问题讨论】:
-
if x: return True \ else: return False - 很高兴我们得到了这些布尔值,嗯?
-
global stringArr并没有按照你的想法去做,你根本不需要那些行 -
你不需要到处处理
KeyboardInterrupt,异常会渗透回顶层,所以就在那里处理吧 -
感谢 gnibbler,我添加了这么多 KeyboardInterrupt,因为如果我只在 main() 中使用它并且脚本例如在 .urlopen 上,它并没有立即关闭,但是所有 KeyboardInterrupt 它确实关闭了. Relet,我不明白你的评论。
-
我相信 relet 是指您在
checkForStr中使用any。在我发布的答案中查看如何简化
标签: python memory memory-leaks