我试图将服务器端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:
/* 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.
}
}
}
这是转换后的代码
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))
我使用simplejson来编码Python列表(因此它将返回一个JSON数组)。我还不能解决这个问题。但是我想我对“回声”做错了什么。
我用这个，效果很好。
from django.utils import simplejson
from django.http import HttpResponse
def some_view(request):
to_json = {
"key1": "value1",
"key2": "value2"
}
return HttpResponse(simplejson.dumps(to_json), mimetype='application/json')
选择:
from django.utils import simplejson
class JsonResponse(HttpResponse):
"""
JSON response
"""
def __init__(self, content, mimetype='application/json', status=None, content_type=None):
super(JsonResponse, self).__init__(
content=simplejson.dumps(content),
mimetype=mimetype,
status=status,
content_type=content_type,
)
在Django 1.7中，JsonResponse对象被添加到Django框架本身，这使得这个任务更加简单:
from django.http import JsonResponse
def some_view(request):
return JsonResponse({"key": "value"})
我通常使用字典而不是列表来返回JSON内容。
import json
from django.http import HttpResponse
response_data = {}
response_data['result'] = 'error'
response_data['message'] = 'Some error message'
在django 1.7之前，你会这样返回它:
return HttpResponse(json.dumps(response_data), content_type="application/json")
对于Django 1.7+，使用JsonResponse，如下所示:
from django.http import JsonResponse
return JsonResponse({'foo':'bar'})
Django代码views.py:
def view(request):
if request.method == 'POST':
print request.body
data = request.body
return HttpResponse(json.dumps(data))
HTML代码view.html:
<!DOCTYPE html>
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#mySelect").change(function(){
selected = $("#mySelect option:selected").text()
$.ajax({
type: 'POST',
dataType: 'json',
contentType: 'application/json; charset=utf-8',
url: '/view/',
data: {
'fruit': selected
},
success: function(result) {
document.write(result)
}
});
});
});
</script>
</head>
<body>
<form>
{{data}}
<br>
Select your favorite fruit:
<select id="mySelect">
<option value="apple" selected >Select fruit</option>
<option value="apple">Apple</option>
<option value="orange">Orange</option>
<option value="pineapple">Pineapple</option>
<option value="banana">Banana</option>
</select>
</form>
</body>
</html>
from django.http import HttpResponse
import json
class JsonResponse(HttpResponse):
def __init__(self, content={}, mimetype=None, status=None,
content_type='application/json'):
super(JsonResponse, self).__init__(json.dumps(content), mimetype=mimetype,
status=status, content_type=content_type)
在视图中:
resp_data = {'my_key': 'my value',}
return JsonResponse(resp_data)
