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Getting a resource in requests: import requests body = requests.get('http://pycurl.io/') Getting a resource with libcurl: import pycurl from
by posix_compliant 8y ago
Getting a resource in requests:
import requests
body = requests.get('http://pycurl.io/')
Getting a resource with libcurl:
import pycurl
from io import BytesIO
buffer = BytesIO()
c = pycurl.Curl()
c.setopt(c.URL, 'http://pycurl.io/')
c.setopt(c.WRITEDATA, buffer)
c.perform()
c.close()
body = buffer.getvalue()
- detaro 8y agoDoesn't answer the question why nobody has build an interface like the first example over the low-level libcurl bindings, or if someone has done so, why it hasn't been more popular. I'd guess: because distributing packages with native dependencies is sort of a a pain, and was way worse in the past, especially cross-platform, and thus python-only packages are preferred. Leaves open why libcurl bindings weren't choosen for stdlib.
- busterarm 8y agoSince we're on the topic of code quality, this example is an incredibly strong indication that you should be using an interface. And while the requests library might be shorter/easier, it doesn't offer nearly the guarantees that it will exhibit the expected behavior nor the feature flexibility that libcurl does. https://curl.haxx.se/libcurl/features.html https://curl.haxx.se/libcurl/features.html https://curl.haxx.se/libcurl/theysay.html https://curl.haxx.se/libcurl/theysay.html