如何让这个程序使用 instagram pic url 并下载？ [复制]

Question

程序的目标是获取用户给定的 instagram url 并允许下载和保存图片。 我已经完成了主要部分，但无法理解如何进一步使用过滤后的正确 url 下载图片并将其保存在我的计算机上。

到目前为止我的代码： 编辑：我添加了一个下载行，但似乎无法获得正确的文件类型？我的意思是它可以保存为我想要的任何内容，但我无法打开它：

import requests
import re
import shutil

def get_response(url):
    r = requests.get(url)
    while r.status_code != 200:
        r.raw.decode_content = True
        r = requests.get(url, stream = True)
    return r.text

def prepare_urls(matches):
    return list({match.replace("\\u0026", "&") for match in matches})

url = input('Enter Instagram URL: ')
response = get_response(url)

vid_matches = re.findall('"video_url":"([^"]+)"', response)
pic_matches = re.findall('"display_url":"([^"]+)"', response)

vid_urls = prepare_urls(vid_matches)
pic_urls = prepare_urls(pic_matches)

if vid_urls:
    print('Detected Videos:\n{0}'.format('\n'.join(vid_urls)))
    print("Can't download video, the provided URL must be of a picture.")
    
if pic_urls:
    print('Detected Pictures:\n{0}'.format('\n'.join(pic_urls)))
        from urllib.request import urlretrieve
        dst = 'Instagram picture.jpg'
        urlretrieve(url, dst)
#EDIT ^

if not (vid_urls or pic_urls):
    print('Could not recognize the media in the provided URL.')
    

Answer 1

我认为这可能会有所帮助...

import requests
from bs4 import BeautifulSoup as bs
import json
import os.path

insta_url = 'https://www.instagram.com'
inta_username = input('enter username of instagram : ')

response = requests.get(f"{insta_url}/{inta_username}/")

if response.ok:
    html = response.text
    bs_html = bs(html, features="lxml")
    bs_html = bs_html.text
    index = bs_html.find('profile_pic_url_hd')+21
    remaining_text = bs_html[index:]
    remaining_text_index = remaining_text.find('requested_by_viewer')-3
    string_url = remaining_text[:remaining_text_index].replace("\\u0026", "&")

    print(string_url, "\ndownloading...")

while True:
    filename = 'pic_ins.jpg'
    file_exists = os.path.isfile(filename)

    if not file_exists:
        with open(filename, 'wb+') as handle:
            response = requests.get(string_url, stream=True)
            if not response.ok:
                print(response)
            for block in response.iter_content(1024):
                if not block:
                    break
                handle.write(block)
    else:
        continue
    break
print("completed")

您可以通过更改文件名变量来更改下载的图像的名称