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Voici le message d'erreur :
Traceback (most recent call last):
File "D:\py\pic_downfrom2255ok.py", line 45, in <module>
html = getHtml(url_all[i])
File "D:\py\pic_downfrom2255ok.py", line 32, in getHtml
html = response.read().decode()
UnicodeDecodeError: 'utf-8' codec can't decode byte 0xb3 in position 184: invalid start byte
Beaucoup de choses ont été modifiées. La raison principale peut être que le site Web cible est codé en gb2312.
Ce programme peut télécharger des images normalement sur d'autres sites Web, mais il y aura des problèmes lors du passage au site Web actuel. quelques conseils. Où est le problème ? J'ai essayé plusieurs méthodes mais rien n'a fonctionné. Le code source est le suivant :
.
#coding=utf-8
import urllib.request
from urllib.request import urlopen, urlretrieve
import urllib
import urllib.parse
import re
import os
from bs4 import BeautifulSoup
url_all =[
'http://www.shop2255.com/showpro/2603.html',
'http://www.shop2255.com/showpro/1558.html',
'http://www.shop2255.com/showpro/1564.html',
'http://www.shop2255.com/showpro/2411.html',
'http://www.shop2255.com/showpro/2409.html',
'http://www.shop2255.com/showpro/1561.html',
'http://www.shop2255.com/showpro/2414.html',
'http://www.shop2255.com/showpro/2609.html',
'http://www.shop2255.com/showpro/2413.html',
'http://www.shop2255.com/showpro/2604.html',
'http://www.shop2255.com/showpro/2605.html',
'http://www.shop2255.com/showpro/2606.html',
'http://www.shop2255.com/showpro/2608.html',
'http://www.shop2255.com/showpro/2607.html',
'http://www.shop2255.com/showpro/2610.html']
def getHtml(url):
response = urlopen(url)
html = response.read().decode("gbk")
return html
def getImg(html):
reg = 'src="(.+?\.jpg)"'
imgre = re.compile(reg)
imglist = re.findall(imgre,html)
return imglist
for i in range(len(url_all)):
html = getHtml(url_all[i])
list=getImg(html.decode())
x = 0
for imgurl in list:
print(x)
file_path = url_all[i]
(filepath,tempfilename) = os.path.split(file_path)
(filename,extension) = os.path.splitext(tempfilename)
if not os.path.exists('d:\%s' % filename):
os.mkdir('d:\%s' % filename)
# os.mkdir('D:\%s' % filename2)
local=r'D:\%s\%s.jpg' % (filename,imgurl.splite("/")[-1])
urllib.request.urlretrieve(imgurl,local)
x+=1
print("done")
天蓬老师2017-05-18 10:55:14
# coding: utf-8
import urllib
import requests
from pyquery import PyQuery as Q
import os
base_url = 'http://www.shop2255.com/'
url_all =['http://www.shop2255.com/showpro/2603.html']
for url in url_all:
_, file_name = os.path.split(url)
dir_name, _ = os.path.splitext(file_name)
if not os.path.exists(dir_name):
os.mkdir(dir_name)
r = requests.get(url)
for _ in Q(r.text).find('img'):
src = Q(_).attr('src')
image_url = src if src.startswith('http') else os.path.join(base_url, src)
_, image_name = os.path.split(image_url)
image_path = os.path.join(dir_name, image_name)
urllib.urlretrieve(image_url, image_path)
漂亮男人2017-05-18 10:55:14
Tout d'abord, dans votre code local=r'D:%s%s.jpg' % (filename,imgurl.splite("/")[-1])
>split s'écrit splite
local=r'D:%s%s.jpg' % (filename,imgurl.splite("/")[-1])
中split
写成了splite
.
还有 urllib.request.urlretrieve(imgurl,local)
这个imgurl
不是一个合法的
url,只是一个相对 url, 要改成绝对 url,需要加上 base_url = 'http://www.shop2255.com/'
.
urllib.request.urlretrieve(imgurl,local)
Ce imgurl
n'est pas une URL légalebase_url = 'http://www.shop2255.com/'
Il semble également y avoir un problème avec le chemin du fichier généré.#🎜🎜#
# -*- coding: utf-8 -*-
import urllib.request
from urllib.request import urlopen, urlretrieve
import urllib
import urllib.parse
import re
import os
from bs4 import BeautifulSoup
base_url = 'http://www.shop2255.com/'
url_all =[
'http://www.shop2255.com/showpro/2603.html',
'http://www.shop2255.com/showpro/1558.html',
'http://www.shop2255.com/showpro/1564.html',
'http://www.shop2255.com/showpro/2411.html',
'http://www.shop2255.com/showpro/2409.html',
'http://www.shop2255.com/showpro/1561.html',
'http://www.shop2255.com/showpro/2414.html',
'http://www.shop2255.com/showpro/2609.html',
'http://www.shop2255.com/showpro/2413.html',
'http://www.shop2255.com/showpro/2604.html',
'http://www.shop2255.com/showpro/2605.html',
'http://www.shop2255.com/showpro/2606.html',
'http://www.shop2255.com/showpro/2608.html',
'http://www.shop2255.com/showpro/2607.html',
'http://www.shop2255.com/showpro/2610.html']
def getHtml(url):
response = urlopen(url)
# print(response.read())
html = response.read().decode("gbk")
print(html)
return html
def getImg(html):
reg = 'src="(.+?\.jpg)"'
imgre = re.compile(reg)
imglist = re.findall(imgre, html)
return imglist
for i in range(len(url_all)):
html = getHtml(url_all[i])
# 注意: 我这里没有你那个错误,我只需要改这个就行了
# list = getImg(html.decode())
list = getImg(html)
# print(list)
x = 0
for imgurl in list:
print(x)
file_path = url_all[i]
(filepath, tempfilename) = os.path.split(file_path)
(filename, extension) = os.path.splitext(tempfilename)
if not os.path.exists('d:\%s' % filename):
os.mkdir('d:\%s' % filename)
# os.mkdir('D:\%s' % filename2)
local = r'D:\%s\%s.jpg' % (filename, imgurl.split("/")[-1])
try:
urllib.request.urlretrieve(base_url + imgurl, local)
except:
print("can't retrieve the" + base_url + imgurl)
x += 1
print("done")