我正在嘗試從 Metacritic 中獲取電影評分。這是拋出錯誤的代碼部分。text = text.replace("_","-")user_agent = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.9.0.7) Gecko/2009021910 Firefox/3.0.7'headers={'User-Agent':user_agent,} URL = "http://metacritic.com/" + textrequest=urllib.request.Request(URL,None,headers)try: response = urllib.request.urlopen(request) data = response.read() soup = BeautifulSoup(data,'html.parser') metacritic_rating = "Metascore: " + soup.find("span",class_="metascore_w").get_text() send_message(metacritic_rating,chat) except: pass我用這個修改了我寫的內(nèi)容：https : //stackoverflow.com/a/42441391/8618880我無法使用，requests.get()因為這個：urllib2.HTTPError: HTTP Error 403: Forbidden我正在尋找一種獲取頁面狀態(tài)代碼的方法。當(dāng)我使用requests.get().我檢查了所有帶有標(biāo)題的答案：urllib.error.HTTPError: HTTP Error 404: Not Found Python但找不到任何幫助。任何幫助表示贊賞。
1 回答
HUX布斯
TA貢獻(xiàn)1876條經(jīng)驗 獲得超6個贊
我想這就是你想要的：
import urllib
user_agent = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.9.0.7) Gecko/2009021910 Firefox/3.0.7'
headers={'User-Agent':user_agent,}
URL = "http://metacritic.com/" + text
request=urllib.request.Request(URL,None,headers)
try:
response = urllib.request.urlopen(request)
data = response.read()
soup = BeautifulSoup(data,'html.parser')
metacritic_rating = "Metascore: " + soup.find("span",class_="metascore_w").get_text()
send_message(metacritic_rating,chat)
except urllib.error.HTTPError as err:
#print(err.code)
if err.code == 403:
<do something>
else:
pass
輸出：
403
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