Created
July 9, 2022 18:33
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#!/usr/bin/env python3 | |
from urllib.request import urlopen | |
from urllib.parse import quote | |
import xml.etree.ElementTree as ET | |
import sys | |
def read_xml(f): | |
it = ET.iterparse(f) | |
for _, el in it: | |
el.tag = el.tag.split('}', 1)[1] # strip all namespaces | |
root = it.root | |
return root | |
key = None | |
page = 0 | |
num_files = 0 | |
if len(sys.argv) == 3: | |
key = sys.argv[2] | |
if len(sys.argv) in (2, 3): | |
base = sys.argv[1] | |
else: | |
print('Usage: python scrapeS3.py BaseURL [StartKey]') | |
sys.exit() | |
out_file = 'urls.txt' | |
with open(out_file, 'a', encoding='utf-8') as f: | |
try: | |
while True: | |
files = list() | |
page += 1 | |
url = f"{base}?marker={quote(key, safe='')}" if key else base | |
sys.stdout.write(f'Enumerating page {page}: {url}\n') | |
sys.stdout.flush() | |
tree = read_xml(urlopen(url)) | |
key = None | |
for entry in tree.findall('Contents'): | |
key = entry.find('Key').text | |
files.append(base + quote(key)) | |
num_files += len(files) | |
f.write('\n'.join(files) + '\n') | |
if key is None: | |
break | |
except KeyboardInterrupt: | |
pass | |
finally: | |
sys.stdout.write('\n') | |
sys.stdout.flush() | |
print(f'Wrote {num_files} URLs to {out_file}!') |
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Is there a way to scrape buckets that show "AccessDenied"?