Python3 get response code from urllib – am I missing something?
Hey folks,
Trying to python3 get response code from urllib but keep getting weird errors or nothing at all. Like, I just wanna check if a URL is alive, ya know?
Using `urllib.request.urlopen()` but the status code feels hidden. Do I really need to dig into `getcode()` or catch exceptions just for this?
```python
import urllib.request
try:
with urllib.request.urlopen('https://example.com') as response:
print(response.getcode()) # finally found it!
except urllib.error.URLError as e:
print("RIP, site's down:", e)
```
Is this the simplest way or am I overcomplicating it?
Also, why doesn’t it just *give* me the code upfront? Feels like extra steps.
Thx in advance!
Yeah, urllib can be a pain for python3 get response code from urllib stuff. Your code’s actually fine, but if you want something simpler, try the `requests` lib.
```python
import requests
r = requests.get('https://example.com')
print(r.status_code)
```
Way cleaner, no nesting. Plus, `requests` handles redirects and timeouts better. Downside? It’s an extra dependency.
Also, urllib hides the code cuz it’s low-level—meant for more control, not convenience.
urllib’s design is just... old school. For python3 get response code from urllib, you’re stuck with `getcode()` or digging into exceptions. Annoying, but that’s how it is.
If you’re just checking URLs, maybe use `curl` in a subprocess?
```python
import subprocess
result = subprocess.run(['curl', '-s', '-o', '/dev/null', '-w', '%{http_code}', 'https://example.com'], capture_output=True, text=True)
print(result.stdout)
```
Overkill? Maybe. Works? Absolutely.
Honestly, your solution’s already the simplest for python3 get response code from urllib without adding libs.
But hey, if you’re doing this a lot, wrap it in a function:
```python
def get_status(url):
try:
with urllib.request.urlopen(url) as resp:
return resp.getcode()
except urllib.error.URLError as e:
return f"Error: {e}"
```
Now you can reuse it and pretend urllib isn’t clunky.
urllib’s like that one friend who makes everything complicated. For python3 get response code from urllib, yeah, `getcode()` is the way.
But if you’re open to alternatives, `httpx` is modern and async-friendly:
```python
import httpx
response = httpx.get('https://example.com')
print(response.status_code)
```
Bonus: it’s faster for bulk checks. Just sayin’.
Wow, thanks y’all! Didn’t expect so many options. Tried `requests` and it’s *so* much easier—why didn’t I switch sooner?
Still weird that python3 get response code from urllib requires so much boilerplate. Might just stick with `requests` unless I’m forced to use stdlib.
Quick Q: anyone know if `httpx` plays nice with proxies? Might need that next.
Your code’s solid for python3 get response code from urllib. The exception handling’s necessary cuz... well, the internet’s unreliable.
If you hate typing, steal this one-liner:
```python
print(urllib.request.urlopen('https://example.com').getcode()) if True else None
```
(Just kidding. Don’t do that. Keep the try/except.)
urllib’s quirks are why everyone uses `requests` for python3 get response code from urllib. But if you’re stuck with stdlib, your approach is correct.
For debugging, add `print(dir(response))`—you’ll see `getcode()` isn’t *that* hidden. Just not obvious.
