Question

模型具有以下关系：

class ShoppingList(models.Model):
   (...)
   config_file = models.FileField(upload_to=upload_config_file)


class FetchedData(models.Model):
    (...)
    config_id = models.ForeignKey(ShoppingList, on_delete=models.CASCADE, default=0)

    def config_link(self):
        return self.config_id.config_file

然后，在serializers.py中：

class FetchedDataSerializer(serializers.ModelSerializer):
    file_link = serializers.SerializerMethodField()

    class Meta:
        model = FetchedData
        fields = ('config_id', 'file_link')

    def get_file_link(self, obj):
        return obj.config_link()

问题是，当我得到响应时，文件被表示为它们的内容，因此，对于JSON文件，我会得到类似的东西：

{
    "config_id": "4544",
    "file_link": [
        "{\r\n",
        "    \"glossary\": {\r\n",
        "        \"title\": \"example glossary\",\r\n",
        "\t\t\"GlossDiv\": {\r\n",
        "            \"title\": \"S\",\r\n",
        "\t\t\t\"GlossList\": {\r\n",
        "                \"GlossEntry\": {\r\n",
        "                    \"ID\": \"SGML\",\r\n",
        "\t\t\t\t\t\"SortAs\": \"SGML\",\r\n",
        "\t\t\t\t\t\"GlossTerm\": \"Standard Generalized Markup Language\",\r\n",
        "\t\t\t\t\t\"Acronym\": \"SGML\",\r\n",
        "\t\t\t\t\t\"Abbrev\": \"ISO 8879:1986\",\r\n",
        "\t\t\t\t\t\"GlossDef\": {\r\n",
        "                        \"para\": \"A meta-markup language, used to create markup languages such as DocBook.\",\r\n",
        "\t\t\t\t\t\t\"GlossSeeAlso\": [\"GML\", \"XML\"]\r\n",
        "                    },\r\n",
        "\t\t\t\t\t\"GlossSee\": \"markup\"\r\n",
        "                }\r\n",
        "            }\r\n",
        "        }\r\n",
        "    }\r\n",
        "}"
    ]
}

该如何仅接收文件位置链接？

Answer 1

在您的config_link方法定义中，您将返回相关的config的{{1}}，它实际上是文件对象本身。

您应该返回文件url，它是文件的file属性：

url

返回文件内容而不是文件位置链接

1 个答案: