如何在WSGI响应中包装普通的JSON内容？

Question

我有一个Django WSGI（不是我的决定）网站，打电话给动态生成JavaScript。我将函数放在views.py中，并且正在接收请求并完成工作，但是返回值正在拒绝。

调用此函数的HTML（网页的JavaScript部分）确实如此：

  var jscript = document.createElement('script');
  jscript.id = 'generate';
  jscript.style.visibility = 'hidden';
  jscript.style.display = 'none';
  jscript.src = `/generate?callback=catchOptions${query}`;  // jsonp https://en.wikipedia.org/wiki/JSONP query is a list of parameters in query string format                                                                                
  if (document.getElementById("generate") == null)
      document.body.appendChild(jscript);  // javascript needs this to work properly             

有映射 /生成到 /generate_planet的地图文件（请参见下文）。进入该功能很棒。这是Djangoff拒绝的回报值。

Here is the function in views.py
from cgitb import reset
from django.shortcuts import render
from . import planetor
from django.http import JsonResponse

def generate_planet(request):
    res = planetor.generate(request.content_params, "/app/planetor/", "FRAMES=1")
    # res is JSON text, NOT a python dict
    return res

# res looks like this:`callback({'camera_location': '-30,-30,-30', 'camera_angle': '30', 'sun_color': '5,5,5', 'sun_position': '10000,0,-10000', 'planet_size': '20.06', 'background': 'background_2.jpg', 'planet': 'surface_1.jpg', 'clouds_size': '1.02', 'clouds': 'clouds_16.jpg', 'clouds_density': '0.80', 'atmosphere': 'iodine', 'atmosphere_density': '0.95', 'atmosphere_size': '1.03', 'moons': '4', 'moon_position': None, 'moon_size': None, 'moon': None, 'random_color': None, 'random_float': None, 'random_trans': None, 'star_system': 'Barnard', 'star_index': 'Zeta', 'planet_index': 'II', 'planet_type': 'Surface ', 'identity': '81654447928', 'designation': 'v_star_index v_star_system v_planet_index', 'clouds_file': 'clouds_16.jpg'})

该功能调用实际上有效，并且“ Planetor.generate（）”运行。问题是，返回JSON（真正的JSONP）确实被Djangoff

Djangoff拒绝了以下：

Internal Server Error: /generate_planet
    Traceback (most recent call last):
      File "/usr/local/lib/python3.9/dist-packages/django/core/handlers/exception.py", line 47, in inner
        response = get_response(request)
      File "/usr/local/lib/python3.9/dist-packages/django/utils/deprecation.py", line 119, in __call__
        response = self.process_response(request, response)
      File "/usr/local/lib/python3.9/dist-packages/django/middleware/clickjacking.py", line 33, in process_response
        response.headers['X-Frame-Options'] = self.get_xframe_options_value(
    AttributeError: 'dict' object has no attribute 'headers'
    [05/Jun/2022 16:52:11] "GET /generate_planet? HTTP/1.1" 500 56694

它正在寻找可以包装在我确定的东西中的返回值，但是我找不到我一生）我可以构建一个和2）任何人使用djangoff这样做任何事情的示例的API文档

Answer 1

我最终弄清楚了。

如果您将请求发送到Django，则是这样：

/my_request?key1=value1&key2=value2&key3=value3

通过任何方式（原始URL，表单提交，Ajax请求等），

在views.py中放置这样的函数。

    def my_request(request):
        selections = request.GET # <== this gets the query string paramaters as a dictionary
        # use the query string parameters as the parameters of the function for creating the answer to the request
        res = {"answer1":"value1","answer2":"value2"} # << python dictionary of answer 
        return JsonResponse(res)  # convert dictionary to JSON

请django捕获该请求并返回JSON答案，请 恢复JSONP，您只需要编码RAW JavaScript：

return 'callback({"answer1":"value1","answer2":"value2"})'