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[英]django-rest-framework: How to serialize join of database models?
[英]How to serialize 2 models with custom relationship in django-rest-framework?
我在数据库中有2个现有表,但我无权更改它们。 显示为以下模型。
class Prenames(models.Model):
typ = models.DecimalField(max_digits=1, decimal_places=0, db_column='tprpretyp')
code = models.DecimalField(max_digits=3, decimal_places=0, db_column='tprprecod')
name = models.CharField(max_length=20, db_column='tprprenam')
class Profiles(models.Model):
userid = models.CharField(max_length=6, db_column='rmsuserid')
prename = models.CharField(max_length=4, db_column='rmsprenam', null=True)
name = models.CharField(max_length=25, db_column='rmsname')
surname = models.CharField(max_length=25, db_column='rmssurnam')
如果SQL我必须
SELECT * FROM Profiles left join Prenames on tprpretyp = int(rmsprenam/1000) and tprprecod = mod(rmsprenam,1000) WHERE trim(rmsuserid) = ?
我已经尝试过的事情:
from rest_framework import serializers
from .models import *
class PrenameSerializer(serializers.ModelSerializer):
class Meta:
model = Prenames
fields = ('type', 'code', 'name')
class ProfileSerializer(serializers.ModelSerializer):
prenames = PrenameSerializer(read_only=True)
class Meta:
model = Profile
fields = ('userid', 'name', 'surname', 'prenames')
我得到的东西:
{
"userid": "560174",
"name": "******",
"surname": "******"
}
我期望的事情:
{
"userid": "560174",
"name": "******",
"surname": "******"
"prenames":[
{
"type:":10,
"code": 01,
"name": "Mr."
}
]
}
我正在使用django-rest-framework 3.2.5和django 1.6如何序列化它们?
尝试这个:
prenames = PrenameSerializer(source="how_you_get_this_field, many=True, read_only=True)
感谢@Aison给我的启发。 这个解决方案不是我想要的。 但这现在可以挽救我的生命。 从SerializerMethodField 。 我决定将ProfileSerializer修改为
class PatientSerializer(serializers.ModelSerializer):
prenameth = serializers.SerializerMethodField('getprenameth')
prenameen = serializers.SerializerMethodField('getprenameen')
def getprenameth(self, obj):
return obj.prename.name
def getprenameen(self, obj):
return obj.prename.en_pre_name
class Meta:
model = Patients
fields = ('userid', 'name', 'surname', 'prenames', 'prenameth', 'prenameen')
这就是我从中得到的。
{
"userid": "560174",
"name": "******",
"surname": "******",
"prenameth": "นาย",
"prenameen": "Mr."
}
现在就足够了。
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