在Python中xpath爬图片src有问题

Question

在Python中xpath爬图片src有问题

在xpath显示正常

Python执行就有问题，src前面有 .. 整个也不一样

尝试换种方式还是不行，直接报错

import urllib.request
from lxml import etree

def create_request(page):
    if(page==1):
        url = 'https://sc.chinaz.com/tupian/fengjing.html'
    else:
        url = 'https://sc.chinaz.com/tupian/fengjing_'+ str(page) + '.html'

    headers = {
        'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/105.0.0.0 Safari/537.36 Edg/105.0.1343.42'
    }
    request = urllib.request.Request(url=url,headers=headers)
    return request

def get_content(request):
    response = urllib.request.urlopen(request)
    content = response.read().decode('utf-8')
    return content

def down_load(content):
    tree = etree.HTML(content)
    name_list = tree.xpath('//div[@class="container"]//div/img/@alt')
    src_list = tree.xpath('//div[@class="container"]//div/img/@src')
    for i in range(len(name_list)):
        name = name_list[i]
        src = src_list[i]
        url = 'https:' + src
        print(name, url)
        # urllib.request.urlretrieve(url=url,filename='./fengjing/' + name + '.jpg')

if __name__ == '__main__':
    start_page = int(input('请输入起始页码'))
    end_page = int(input('请输入结束页'))

    for page in range(start_page,end_page+1):
        request = create_request(page)
        content = get_content(request)
        down_load(content)

Answer 1

import urllib.request
from lxml import etree


def create_request(page):
    if (page == 1):
        url = 'https://sc.chinaz.com/tupian/fengjing.html'
    else:
        url = 'https://sc.chinaz.com/tupian/fengjing_' + str(page) + '.html'
    ##请求头中我添加Host和Refer参数，也是防止反爬的一种手段
    headers = {
        'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/105.0.0.0 Safari/537.36 Edg/105.0.1343.42',
        'Host':'sc.chinaz.com',
        'Refer':'https://sc.chinaz.com/tupian/fengjing.html'
    }
    request = urllib.request.Request(url=url, headers=headers)
    return request

def get_content(request):
    response = urllib.request.urlopen(request)
    content = response.read().decode('utf-8')
    return content


def down_load(content):
    tree = etree.HTML(content)
    name_list = tree.xpath('//div[@class="container"]//div[contains(@class,"item")]//img')
    # 我试过用 @src 属性获取值，但是和我想要结果不一样，所以下面就用 @data-original 获取图片的url地址
    for item in name_list:
        img_name = item.xpath( './@alt' )[0]
        img_url ='https:' +  item.xpath( './@data-original' )[0]
        print(img_name,img_url)
        urllib.request.urlretrieve(url=img_url, filename='./fengjing/' + img_name + '.jpg')


if __name__ == '__main__':
    start_page = int(input('请输入起始页码'))
    end_page = int(input('请输入结束页'))

    for page in range(start_page, end_page + 1):
        request = create_request(page)
        content = get_content(request)
        down_load(content)