Question

我无法让Django运行带参数的Tweepy脚本。当我启动服务器并创建一个实例时，我似乎能够让脚本运行print语句，但是我很难让它从HTML输入中获取标记。

HTML按钮：

<div class="col s12">
      <div class="input-field inline">
      <label for="User">Enter a term</label>
        <form action = "submit/" method = "post">
        {% csrf_token %}
       <input id="Country" type="text" name="Country" maxlength="100" required value="{{ tag }}" />
       <button class="btn waves-effect red" type="submit" name="action">Submit
        <i class="material-icons right">send</i>
      </button>
       </form>
      </div>

Views.py

from django.shortcuts import render
from .forms import GetTweets
from . import live

# Create your views here.

#Index by Kenny P
def index(request):
    context = {'tag': ''}
    return render(request, 'visualize/index.html', context)

#Tweets by Kenny P. Tweet will return the tweets
def Tweets(request):
    #Terms we want displayed go here!
    print 'IN!!!'

    if request.method == 'POST':
        info = GetTweets(request.POST)

        if info.is_valid():
            country = request.POST['Country']
            tw = live.GetTweets(render_data.cleaned_data['Country'])

        else:
            country =  GetTweets()

    context = {
        'Country': country,
    }

    return render(request, 'visualize/index.html', context)

def submit(request):
    return render(request, 'visualize/index.html')

forms.py

from django import forms

#Search for the tag
class GetTweets(forms.Form):
    user = forms.CharField(label='User', max_length=30)

最后，（临时）tweepy脚本：

import tweepy
import time
from collections import defaultdict
import json

consumer_key = "??"
consumer_secret = "??"
access_key = "??"
access_secret = "??"

auth = tweepy.OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_key, access_secret)
api = tweepy.API(auth)


def GetTweets(tag):
  placeholder = []
  for tweet in tweepy.Cursor(api.search, q=tag).items(20):
    placeholder.append(tweet.created_at)
    placeholder.append(tweet.text)
  #created_at
  print '   placeholder 0 \n'
  print placeholder[0] 
  print placeholder[1]
  print ' placeholder 1\n'
  print placeholder[2]
  print placeholder[3]

URLS

from django.conf.urls import url
from django.contrib import admin
from . import views
urlpatterns = [
    url(r'^$', views.index, name='index'),
    url(r'^submit', views.Tweets),
]

Answer 1

您必须将listView.setOnGroupClickListener(mListener);传递给您的模板，否则tag将为空。更改您的{{ tag }}视图（iirc，在旧版本的Django中，您需要在index()中包含上下文字典：

template.Context(context)

将您的html更改为发布到正确的网址，例如：

context = {'tag': 'hello world']
return render(request, 'visualize/index.html', context)

确保urls.py匹配（您最后缺少<form action="submit/" method="post">）：

/$

你的urls.py也需要匹配视图定义。如果您将视图定义为采用url(r'^submit/$', views.Tweets),参数：

tag

然后urls.py需要捕获名为def Tweets(request, tag):的字段，例如：

tag

但是，由于您将url(r'^submit/(?P<tag>\w+)/$', views.Tweets),变量作为Country变量传递，因此您的网址应为

POST

并且应该在没有标记参数的情况下定义视图

url(r'^submit/$', views.Tweets),

让Django运行带参数的脚本？

1 个答案: