爬虫图片 - 请教各位：python爬虫编码问题，版本3.6，win10 64位下？

问题描述

问 题

这是报错信息：

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

改了好多地方，主要可能是目标网站是gb2312编码，
这个程序在别的网站是可以正常下载图片的，换上现在的网站就有问题
还请各位多多指教，问题出在哪里？试了几个方法都不行
源码如下：

#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")

解决方案

# 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)

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