2015-09-04 100 views
0

我要救JSON在Django场像以下:字符串保存为JSON在Django

{ 
    "first_name": "bgbg", 
    "last_name": "hbbgh", 
    "username": "reza", 
    "user_profile": { 
     "id": "1ac3ca9c-563e-4f78-b9fa-01552a30585d", 
     "gender": "M", 
     "birthday": "2015-09-04", 
     "country": "Iran", 
     "city": "Tehran", 
     "street_address": "dsafrfl", 
     "state": "dskj", 
     "about": "fdgrdshrbfr", 
     "social_links": {"linkedin": "http://linkedin.com"}, 
     "location": "11,22", 
     "avatar": "http://127.0.0.1:8000/media/user/1ac3ca9c-563e-4f78-b9fa-01552a30585d/avatar/1ac3ca9c-563e-4f78-b9fa-01552a30585dBooking.gif", 
     "verification_code": "9469" 
    } 

但social_links保存如下:

"social_links": "{\"linkedin\": \"http://linkedin.com\"}", 

我使用Django的REST框架

class Profile(models.Model): 
    id = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False) 
    avatar = models.ImageField(blank=True, null=True, upload_to=get_image_path) 
    city = models.CharField(max_length=100, blank=True) 
    state = models.CharField(max_length=100, blank=True) 
    social_links = JSONCharField(max_length=200, blank=True) 
... 
+1

您确定您不只是在调试器中查看未转义的字符串吗? – Sayse

+0

我不知道你说的究竟是什么意思? – altruistic

+1

你可以保存为文本字段并返回响应作为返回响应(json.loads(social_links),status = status.HTTP_200_OK) –

回答

0

你应该解析你json数据然后用(objects.create)命令保存。
你也可以使用“Django rest-framework”序列化器。