我怎麼可以把輸入取（（URL））這個函數

Question

import random 
import urllib.request 

def down_load_imag(url): 
    name = random.randrange(1, 1000) 
    full_name = str(name) + ".jpg" 
    urllib.request.urlretrieve(url, full_name) 

down_load_imag ("http://3.bp.blogspot.com/-WjQkpjkw9uQ/Vij8lG0pCdI/AAAAAAAAAJ4/-CifLZ5KG-Y/s1600/fedora_infinity_140x140.png") 

Answer 1

在python3必須使用input()功能：

import random 
import urllib.request 

def down_load_imag(url): 
    name = random.randrange(1, 1000) 
    full_name = str(name) + ".jpg" 
    urllib.request.urlretrieve(url, full_name) 
url = input('Enter url: ') 
# here you can evaluate url with regex and if passed: 
down_load_imag (url) 

Answer 2

Python帶有的argparse模塊。假設你的代碼生活downloader.py，再補充一點：

import argparse 
import sys 

parser = argparse.ArgumentParser(description="Download image to random filename") 
parser.add_argument("--url", help="image url to download") 
args = parser.parse_args() # parses sys.argv 

url = args.url 

if url is None: 
    print("Give me an URL please!", file=sys.stderr) 
    sys.exit(-1) 

down_load_imag(url) 

然後，使用這樣的：

python3 downloader.py --url http://what.ever.com/image.jpg 

您可以優化該功能拍攝使用input()如果url是無的URL。