我將 1000 多個 URL 的列表（這些 URL 用于下載報告）保存在一個.csv文件中。有些 URL 已經存在404 error，我想找到一種方法將它們從列表中刪除。我設法編寫了一段代碼來識別下面哪個 URL 無效（對于 python 3）。但是，由于存在許多 URL，我不知道如何自動從列表中刪除這些 URL。謝謝你！from urllib.request import urlopenfrom urllib.error import HTTPErrortry: urlopen("url")except HTTPError as err: if err.code == 404: print ('invalid') else: raise
4 回答
犯罪嫌疑人X
TA貢獻2080條經驗 獲得超4個贊
您可以使用另一個列表來保存404 url（如果404 url小于正常url），然后獲取差異集，所以：
from urllib.request import urlopen
from urllib.error import HTTPError
exclude_urls = set()
try:
urlopen("url")
except HTTPError as err:
if err.code == 404:
exclude_urls.add(url)
valid_urls = set(all_urls) - exclude_urls
至尊寶的傳說
TA貢獻1789條經驗 獲得超10個贊
你可以這樣做：
from urllib.request import urlopen
from urllib.error import HTTPError
def load_data(csv_name):
...
def save_data(data,csv_name):
...
links=load_data(csv_name)
new_links=set()
for i in links:
try:
urlopen("url")
except HTTPError as err:
if err.code == 404:
print ('invalid')
else:
new_links.add(i)
save_data( list(new_links),csv_name)
慕娘9325324
TA貢獻1783條經驗 獲得超5個贊
嘗試這樣的事情：
from urllib.request import urlopen
from urllib.error import HTTPError
# 1. Load the CSV file into a list
with open('urls.csv', 'r') as file:
reader = csv.reader(file)
urls = [row[0] for row in reader] # Assuming each row has one URL
# 2. Check each URL for validity using your code
valid_urls = []
for url in urls:
try:
urlopen(url)
valid_urls.append(url)
except HTTPError as err:
if err.code == 404:
print(f'Invalid URL: {url}')
else:
raise # If it's another type of error, raise it so you're aware
# 3. Write the cleaned list back to the CSV file
with open('cleaned_urls.csv', 'w') as file:
writer = csv.writer(file)
for url in valid_urls:
writer.writerow([url])
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