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我有这段代码可以在 python 2 中读取 Mashape.com API。我如何在 python 3 中读取它?

代码

import urllib, urllib2, json
from pprint import pprint

URL = "https://getsentiment.p.mashape.com/"
text = "The food was great, but the service was slow."
params = {'text': text, 'domain': 'retail', 'terms': 1, 'categories': 1,'sentiment': 1, 'annotate': 1}
headers = {'X-Mashape-Key': YOUR_MASHAPE_KEY}

opener = urllib2.build_opener(urllib2.HTTPHandler)
request = urllib2.Request(URL, urllib.urlencode(params), headers=headers)
response = opener.open(request)
opener.close()

data = json.loads(response.read())
pprint(data)

我试过这段代码,但它有以下错误:

import urllib.parse
import urllib.request

URL = "https://getsentiment.p.mashape.com/"
text = "The food was great, but the service was slow."
params = {'text': text, 'domain': 'retail', 'terms': 1, 'categories': 1, 'sentiment': 1, 'annotate': 1}
headers = {'X-Mashape-Key': YOUR_MASHAPE_KEY}

opener = urllib.request.build_opener(urllib.request.HTTPHandler)
request = urllib.request.Request(URL, urllib.parse.urlencode(params), headers)
response = opener.open(request)
opener.close()

data = json.loads(response.read())
print(data)

错误 :

TypeError: POST data should be bytes or an iterable of bytes. It cannot be of type str.
4

1 回答 1

1

在这一行:

request = urllib.request.Request(URL, urllib.parse.urlencode(params), headers)

尝试替换为

data = urllib.parse.urlencode(params).encode('utf-8')
request = urllib.request.Request(URL, data, headers)
于 2016-09-13T15:43:48.770 回答