Question

此代码应该询问2个可选参数和1个必需参数，'format'和'type'参数是可选的，'url'是必需参数，它将是您要从中下载图像的网站地址（jpg或/和.png类型图像，取决于你在'type'参数中给出的值，并将其保存为单个图像或单个JSON文件（取决于你在'format'参数中给出的值），我得到错误你可以告诉我哪里出错了，这是我的代码..

import argparse
import base64
import json
import os
from bs4 import BeautifulSoup
import requests



def scrape(url,format_,type_):
    try:
        page = requests.get(url)
    except requests.RequestException as rex:
        print(str(rex))
    else:
        soup = BeautifulSoup(url,'html.parser')
        images = _fetch_images(soup,url)
        images = _filter_images(images,type_)
        _save(images,format_)

def _fetch_images(soup,base_url):
    images = []
    for img in soup.find_all('img'):
        src = img.get('src')
        img_url = ('{base_url}/{src}'.format(base_url,src))
        name = img_url.split('\\')[-1]
        images.append(dict(name=name,url=img_url))
return images

def _filter_images(images,type_):
    if type == 'all':
        return images
    ext_map = {'png':['.png'],'jpg':['.jpg','.jpeg']}
    return [img for img in images if 
            _match_extension(img['name'],ext_map(type_))]

def _match_extension(filename,extension_list):
    name,extension = os.path.splittext(filename.lower())
    return extension in extension_list

def _save(images,format_):
    if images:
        if format_=='img':
            _save_images(images)
        else:
            _save_json(images)
            print('Done!')
    else:
        print('there are no images!')

def _save_images(images):
    for img in images:
        img_data = requests.get(img['url']).content
        with open(img['name'],'wb') as f:
        f.write(img_data)

def _save_json(images):
    data = {}
    for img in images:
        img_data = requests.get(img['url']).content
        b64_img_data = base64.b64encode(img_data)
        str_img_data = b64_img_data.decode('utf-8')
        data[img['name']]=str_img_data
    with open('images.json','w') as ijson:
        ijson.write(json.dump(data))


if __name__ == "__main__":
    parser = argparse.ArgumentParser(
        description='Scrape a webpage.')
    parser.add_argument(
        '-t',
        '--type',
        choices=['all', 'png', 'jpg'],
        default='all',
        help='The image type we want to scrape.')
    parser.add_argument(
        '-f',
        '--format',
        choices=['img', 'json'],
        default='img',
        help='The format images are saved to.')
    parser.add_argument(
        'url',
        help='The URL we want to scrape for images.')
    args = parser.parse_args()
    args.url = str(input('enter the url\n'))
    args.format = str(input('enter img or json \n'))
    args.type = str(input('enter all or png or jpg'))
    scrape(args.url,args.format,args.type)

我想在PyCharm中运行它，我得到的错误是，

usage: scrape.py [-h] [-t {all,png,jpg}] [-f {img,json}] url
scrape.py: error: the following arguments are required: url

Answer 1

为了在Pycharm中将参数传递给您的函数，请导航到Run -> Edit Configurations并在那里设置参数。

或者，尝试使用命令行指定的参数调用程序。

如何在python中传递argparse参数中的输入

1 个答案: