尝试打开网址时，我在使用Mechanize时遇到406错误：
for url in urls:
if "http://" not in url:
url = "http://" + url
print url
try:
page = mech.open("%s" % url)
except urllib2.HTTPError, e:
print "there was an error opening the URL, logging it"
print e.code
logfile = open ("log/urlopenlog.txt", "a")
logfile.write(url + "," + "couldn't open this page" + "\n")
continue
else:
print "opening this URL..."
page = mech.open(url)
知道什么会导致406错误发生？如果我转到相关网址，我可以在浏览器中打开它。
答案 0 :(得分：2)
尝试根据浏览器发送的内容为您的请求添加标头;首先添加Accept标题（406通常意味着服务器不喜欢您想要接受的内容）。
请参阅文档中的"Adding headers"：
req = mechanize.Request(url)
req.add_header('Accept', 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8')
page = mechanize.urlopen(req)
Accept标头值取决于Chrome发送的标头。
答案 1 :(得分：0)
如果您想了解浏览器发送的标题，此网页会向您显示：https://www.whatismybrowser.com/detect/what-http-headers-is-my-browser-sending
接受＆＃39;和＆＃39;用户代理＆＃39;标题应该足够了。这就是我为摆脱错误所做的工作：
#establish counter
j = 0
#Create headers for webpage
headers = {'User-Agent': 'Mozilla/5.0', 'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8'}
#Create for loop to get through list of URLs
for url in URLs:
#Verify scraper agent so that web security systems don't block webpage scraping upon URL opening, with j as a counter
req = mechanize.Request(URLs[j], headers = headers)
#Open the url
page = mechanize.urlopen(req)
#increase counter
j += 1
你也可以尝试导入＆＃34; urllib2＆＃34;或＆＃34; urllib＆＃34;用于打开这些URL的库。语法是一样的。