問題描述
需要幫助從 curl 遷移到 pycurl (need help with moving from curl to pycurl)
I'm trying to figure out how to send a pycurl "POST" to a REST interface (in this case, Neo4j).
I can do it via CURL with:
curl ‑HContent‑Type:application/json ‑X POST ‑d '"http://localhost:7474/db/data/node/123"' http://localhost:7474/db/data/index/node/persons/name/peter
... but I cannot figure out how to translate that with CURL options. This is the closest that I've gotten but it fails
data = "http://localhost:7474/db/data/node/123"
headers = {"Content‑Type": "application/json"}
url_dir = "http://localhost:7474/db/data/index/node/persons/name/peter"
conn = pycurl.Curl()
c.setopt(pycurl.URL, url_dir)
c.setopt(pycurl.HTTPHEADER, headers)
what do I do with the data to transmit?
edit: by data I mean the variable entitled data. How do I send the data (what's included in the ‑d option) via pycurl. I've tried urlencode‑ing it but get an error (obviously, because of its type). I've tried urllib.quote_plus() but that's caused a 400 error on the REST service. I suspect that this is the issue since my GETs are working just fine.
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參考解法
方法 1:
Appending these statements may help:
c.setopt(pycurl.POST, 1)
c.setopt(pycurl.POSTFIELDS, data)
Check out the "HTTP POSTing" section of the libcurl tutorial for more info.
Note 1: If the data string must be enclosed by double quotes, then you should do: c.setopt(pycurl.POSTFIELDS, '"%s"' % data).
Note 2: I think you made a typo: the conn variable must be called c.