直接生成Django REST API?

看来在django没有办法这样做。

这不是最好的解决方案,

有时,根据关系需要不同的序列化程序类

我刚刚创建了一个有效的脚本:

您只需要从项目目录运行它,并将应用程序名称作为参数

`import os
import sys

ROUTER_TEMPLATE = "router.register(r'{}', views.{}ViewSet)"
SERIALIZER_TEMPLATE = """
class {}Serializer(serializers.HyperlinkedModelSerializer):

    class Meta:
        model = {}
        fields = '__all__'
"""
VIEW_TEMPLATE= """

class {}ViewSet(viewsets.ModelViewSet):
    queryset = {}.objects.all()
    serializer_class = {}Serializer


"""

def gen_api(app_name):
        models= open(app_name+'/models.py', 'r')
    aLines = models.readlines()

    for sLine in aLines:

        aLine = sLine.split()

        for sWord in aLine:

            if sWord == "class" and aLine[aLine.index(sWord)+1] != "Meta:":
                model_name= aLine[aLine.index(sWord)+1].split("(")[0]
                gen(app_name,model_name )




def gen(app_name, name):

    config = open(app_name+'/urls.py', 'a') #append
    config.write("\n"+ ROUTER_TEMPLATE.format(name, name))
    config.close()

    config = open(app_name+'/serializers.py', 'a') #append
    config.write("\n"+ SERIALIZER_TEMPLATE.format(name, name))
    config.close()

    config = open(app_name+'/views.py', 'a') #append
    config.write("\n"+ VIEW_TEMPLATE.format(name, name, name))
    config.close()

if __name__ == "__main__":

    gen_api(sys.argv[1])
`