suchen

Heim  >  Fragen und Antworten  >  Hauptteil

Crawler-Bilder – Bitte sagen Sie mir: Python-Crawler-Codierungsproblem, Version 3.6, Win10 64-Bit?

Das ist die Fehlermeldung:

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

Der Hauptgrund dafür ist möglicherweise, dass die Zielwebsite in GB2312 codiert ist.
Dieses Programm kann Bilder normalerweise auf andere Websites herunterladen. Beim Wechsel zur aktuellen Website wird es jedoch Probleme geben ein paar Ratschläge. Wo liegt das Problem? Ich habe mehrere Methoden ausprobiert, aber nichts hat funktioniert. Der Quellcode lautet wie folgt:

#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")
伊谢尔伦伊谢尔伦2750 Tage vor975

Antworte allen(2)Ich werde antworten

  • 天蓬老师

    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)

    Antwort
    0
  • 漂亮男人

    漂亮男人2017-05-18 10:55:14

    首先在你这个代码里面 local=r'D:\%s\%s.jpg' % (filename,imgurl.splite("/")[-1])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/'

    还有生成的文件路径好像也有问题.

    # -*- 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")
    

    Antwort
    0
  • StornierenAntwort