代码之家 › 专栏 › 技术社区 › lalit bhadouria

如何使用django framework发送图像文件而不保存在数据库中

django-views django-templates django-forms django

0

lalit bhadouria · 技术社区 · 8 年前

我是Django的新手,我被Django的观点困住了。py am表单中的两个字段。py 1st CharField 2nd ImageField我的CharField已成功渲染,但ImageField未渲染如何解决此问题我想用tweet发送图像文件,而不保存在数据库中请帮助我解决此问题谢谢

我的文件在这里-

形式。py公司

from django import forms

class HomeForm(forms.Form):
  tweet = forms.CharField()
  img = forms.ImageField(label='Select a Image file',)

意见。py公司

from django.http import HttpResponseRedirect
from django.shortcuts import render
from .forms import *
from django.views.generic import TemplateView


class Texting(TemplateView):        
  Template_name =  'app_temp/xyz.html'

  def get(self, request):
    form = HomeForm()
    return render(request, self.Template_name, {'form': form})

  def post(self, request):
    access_token = 'XYZ'
    access_token_secret = 'PQR'
    consumer_key = 'LMN'
    consumer_secret = 'ABC'
    api = TwitterAPI(consumer_key, consumer_secret , access_token , access_token_secret )
    form = HomeForm(request.POST , request.FILES)
    if form.is_valid():
        text = form.cleaned_data['tweet']
        img = form.request.FILES['img'] 

        r = api.request('statuses/update_with_media', {'status':text}, {'media[]':img})
    return render(request,'app_temp/new_page.html')

html

{% extends 'login_base.html'%}

{% block content %}

<div class="container">
   <form method="POST">
         {% csrf_token %}
         {{ form.as_p }}
         <button type="submit" class="btn btn-primary">Tweet</button>
   </form>
   <h1>{{img}}</h1>
    <h1>{{text}}</h1>
 </div>
{% endblock  %}

2 回复 | 直到 7 年前

1

0

Jonas 8 年前

statuses/update_with_media 已弃用。此API调用已被两步过程所取代。首先,使用上载图像 media/upload . 其次,使用媒体ID发布您的推文 statuses/update . 您可以在以下位置找到完整的示例: https://github.com/geduldig/TwitterAPI/blob/master/examples/update_media.py .

# STEP 1 - upload image
file = open(IMAGE_PATH, 'rb')
data = file.read()
r = api.request('media/upload', None, {'media': data})
print('UPLOAD MEDIA SUCCESS' if r.status_code == 200 else 'UPLOAD MEDIA FAILURE')

# STEP 2 - post tweet with a reference to uploaded image
if r.status_code == 200:
    media_id = r.json()['media_id']
    r = api.request(
        'statuses/update', {'status': TWEET_TEXT, 'media_ids': media_id})
    print('UPDATE STATUS SUCCESS' if r.status_code == 200 else 'UPDATE STATUS FAILURE')

2

0

lalit bhadouria 8 年前

形式。py公司

from django import forms
from app.models import *

class HomeForm(forms.Form , forms.ModelForm):
  tweet = forms.CharField(max_length = 280 ,widget = forms.Textarea)
  image = forms.ImageField(label='Select a Image file',)

意见。py公司

from django.shortcuts import render
from django.views.generic import TemplateView
from TwitterAPI import *
from app.forms import *
class Texting(TemplateView):        
   def post(self, request):
      form = HomeForm(request.POST,request.FILES)
      if form.is_valid():
        text = form.cleaned_data['tweet']
        image = form.cleaned_data['image']