requests the url, get the http response code , if the code is 200, open the url on browser

Author: JavaProgrammerLB

findsth.py

@@ -0,0 +1,37 @@
+import webbrowser
+import sys
+import requests
+import logging
+
+protocol='http'
+topDomainName=['.com','.cn','.net','.de','.fi','.co.jp', '.ru']
+argv=sys.argv[1:]
+print(argv)
+
+logging.basicConfig(level=logging.INFO)
+urls=[]
+for i in range(len(topDomainName)):
+	aTop=topDomainName[i]
+	for j in range(len(argv)):
+		domainName=argv[j]
+		url=protocol + '://' + 'www.' + domainName + aTop
+		logging.info('url is: %s',url)
+		try:
+			req=requests.get(url, timeout=10)
+			code=req.status_code
+		except requests.exceptions.ReadTimeout:
+			code=-1
+		except requests.exceptions.ConnectionError:
+			code=-1
+		except requests.exceptions.TooManyRedirects:
+			code=-1
+		logging.info('code is: %d',code) 
+		if code == 200:
+			urls.append(url)
+			logging.info('Available url:%s',url)
+
+for i in range(len(urls)):
+	logging.info('Openning: %s',urls[i])
+	webbrowser.open_new_tab(urls[i])
+#webbrowser.open_new_tab(url)
+