python - 如何向 Thomson Reuters Open Calais API 提交 Python 请求

Question

我在 Google App Engine (gae) 上有一些 Python 代码，它向 Thomson Reuters Open Calais API 发出请求:它发送一些纯文本作为数据，并且应该作为返回获得一个带有与文本对应的标签的 JSON 对象。不幸的是，我得到的只是错误 500 以及 gae 控制台中的以下错误消息:

上述错误似乎发生在 Werkzeug 中，它是一个 Python WSGI 实用程序库(无论 WSGI 是什么意思!)。错误出现在第 117 行 here ，所以这可能意味着缺少字符集，但我不知道应该在我的 Python 代码中的哪个位置设置字符集。

如果有人能帮我解决这个问题，我将不胜感激。

这是我的 Python 代码:

from google.appengine.api import urlfetch
from flask import Flask, request

app = Flask(__name__)

calais_url = "https://api.thomsonreuters.com/permid/calais"

@app.route('/', methods=['POST'])
def main():
    data = "The 'In' camp holds a one-point lead ahead of Britain's June 23 referendum on whether the country should remain in the European Union, according to an online opinion poll by YouGov released on Friday.The poll found support for 'In' stood at 40 percent, while 39 percent intended to vote 'Out', 16 percent were undecided and 5 percent did not intend to vote.The poll of 3,371 people was conducted between April 12 and 14 and the results were similar to those seen in other recent YouGov polls. The previous one, conducted on April 11-12, found 'In' and 'Out' were tied on 39 percent with 17 percent undecided."
    headers = {'X-AG-Access-Token' : 'my_secret_key', 'Content-Type' : 'text/raw', 'outputFormat' : 'application/json', 'omitOutputtingOriginalText' : 'true'}
    try:
        sendArticleText(data, headers)
    except Exception ,e:
        return 'Error in connect ' , e

def sendArticleText(_data, _headers):
    response = urlfetch.fetch(calais_url, payload=_data , method=POST, headers=_headers, deadline=60)
    content = response.text
    if response.status_code == 200:
        return content.json()

@app.errorhandler(404)
def page_not_found(e):
    """Return a custom 404 error."""
    return 'Sorry, Nothing at this URL.', 404

@app.errorhandler(500)
def page_not_found(e):
    """Return a custom 500 error."""
    return 'Sorry, unexpected error:\n{}'.format(e), 500

提前谢谢你。

Answer 1

发生这种情况是因为您从 main 返回一个元组，如果 View 函数返回一个元组，flask 期望元组中的第二个成员是 http 状态代码，而不是异常值。

代替

return 'Error in connect ' , e

第 15 行应该是这样的:

return 'Error in connect: {}'.format(e)

这将使您看到真正的错误。