使用Django和Python创建JSON响应 [英] Creating a JSON response using Django and Python

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本文介绍了使用Django和Python创建JSON响应的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在尝试将服务器端的Ajax响应脚本转换为Django HttpResponse,但显然它不起作用。



这是服务器端脚本: / p>

  / *接收值* / 
$ validateValue = $ _ POST ['validateValue'];
$ validateId = $ _ POST ['validateId'];
$ validateError = $ _ POST ['validateError'];

/ *返回值* /
$ arrayToJs = array();
$ arrayToJs [0] = $ validateId;
$ arrayToJs [1] = $ validateError;

if($ validateValue ==Testuser){//验证?
$ arrayToJs [2] =true; // RETURN TRUE
echo'{jsonValidateReturn:'。json_encode($ arrayToJs)。'}'; //返回阵列成功
}
else {
($ x = 0; $ x <1000000; $ x ++){
if($ x == 990000){
$ arrayToJs [2] =false;
echo'{jsonValidateReturn:'。json_encode($ arrayToJs)。'}'; // RETURBS ARRAY WITH ERROR。
}
}
}

这是转换后的代码

  def validate_user(request):
如果request.method =='POST':
vld_value = request.POST.get('validateValue')
vld_id = request.POST.get('validateId')
vld_error = request.POST.get('validateError')

array_to_js = [vld_id,vld_error,False]

如果vld_value ==TestUser:
array_to_js [2] = True
x = simplejson.dumps(array_to_js)
返回HttpResponse(x)
else:
array_to_js [2] = False
x = simplejson.dumps(array_to_js)
error ='Error'
return render_to_response('index .html',{'error':error},context_instance = RequestContext(request))
return render_to_response('index.html',context_instance = RequestContext(request))

我使用simplejson对Python列表进行编码(因此它将返回一个JSON数组)。我找不到问题。但是我觉得我对'echo'做了一些错误。

解决方案

我通常使用一个字典而不是列表返回JSON内容

  import json 

from django.http import HttpResponse

response_data = {}
response_data ['result'] ='error'
response_data ['message'] ='一些错误消息'

前Django 1.7你会这样返回:

  return HttpResponse json.dumps(response_data),content_type =application / json)

对于Django 1.7+,使用 JsonResponse ,如此SO答案所示:



django.http import JsonResponse
return JsonResponse({'foo':'bar'})
 $ p> 

I'm trying to convert a server side Ajax response script into a Django HttpResponse, but apparently it's not working.

This is the server-side script:

/* RECEIVE VALUE */
$validateValue=$_POST['validateValue'];
$validateId=$_POST['validateId'];
$validateError=$_POST['validateError'];

/* RETURN VALUE */
$arrayToJs = array();
$arrayToJs[0] = $validateId;
$arrayToJs[1] = $validateError;

if($validateValue =="Testuser"){  // Validate??
    $arrayToJs[2] = "true";       // RETURN TRUE
    echo '{"jsonValidateReturn":'.json_encode($arrayToJs).'}';  // RETURN ARRAY WITH success
}
else{
    for($x=0;$x<1000000;$x++){
        if($x == 990000){
            $arrayToJs[2] = "false";
            echo '{"jsonValidateReturn":'.json_encode($arrayToJs).'}';   // RETURNS ARRAY WITH ERROR.
        }
    }
}

And this is the converted code

def validate_user(request):
    if request.method == 'POST':
        vld_value = request.POST.get('validateValue')
        vld_id = request.POST.get('validateId')
        vld_error = request.POST.get('validateError')

        array_to_js = [vld_id, vld_error, False]

        if vld_value == "TestUser":
            array_to_js[2] = True
            x = simplejson.dumps(array_to_js)
            return HttpResponse(x)
        else:
            array_to_js[2] = False
            x = simplejson.dumps(array_to_js)
            error = 'Error'
            return render_to_response('index.html',{'error':error},context_instance=RequestContext(request))
    return render_to_response('index.html',context_instance=RequestContext(request))

I'm using simplejson to encode the Python list (so it will return a JSON array). I couldn't figure out the problem yet. But I think that I did something wrong about the 'echo'.

解决方案

I usually use a dictionary, not a list to return JSON content.

import json

from django.http import HttpResponse

response_data = {}
response_data['result'] = 'error'
response_data['message'] = 'Some error message'

Pre-Django 1.7 you'd return it like this:

return HttpResponse(json.dumps(response_data), content_type="application/json")

For Django 1.7+, use JsonResponse as shown in this SO answer like so :

from django.http import JsonResponse
return JsonResponse({'foo':'bar'})

这篇关于使用Django和Python创建JSON响应的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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