如何在Django REST中将字符串添加到ModelSerializer [英] How to add string to a ModelSerializer in Django REST

问题描述

在我的数据库中，我存储了特定图像的文件名。假设这是来自django的 models.py

In my database, I store the file name of a particular image for an item. Let's say this is the model in models.py

from django.db import models

class Product(models.Model):
    sku = models.CharField(validators=[isalphanumeric], max_length=20, null=False, blank=False)
    image = models.CharField(max_length=20, blank=False, null=False)

我有一个像这样定义的序列化器，在 serializers.py

and then I have a serializer defined like so in serializers.py

from rest_framework import serializers
from app.models import Product

class ProductSerializer(serializer.ModelSerializer):
    class Meta:
        model = Product
        fields = '__all__'

我想要的是能够将字符串添加到<$ c $的图像中c> Product 使其成为表示相对链接的字符串，例如：

what I want is to be able to add a string to the image of a Product that makes it into a string representing the relative link, something like:

storage_location = '/assets/img'
img_url = f'{storage_location}/{image}'

之所以要这样做，是因为我想灵活使用url，而不是将文件名设置为文件位置，然后每次我都必须更新数据库更改我排列图像的方式（我仍然不确定如何存储它们）。

The reason why I want to do this is because I want to be flexible with the urls rather than having the file name be a "file location" and then having to update the database each time I change how I arrange my images (I'm still not sure how to store them).

我该怎么做？

推荐答案

首先，您可以使用模型的 ImageField 为此：

First of all you can use model's ImageField for this:

class Product(models.Model):
    sku = models.CharField(validators=[isalphanumeric], max_length=20, null=False, blank=False)
    image = models.ImageField(max_length=20, blank=False)

这将自动添加 MEDIA_URL 设置为获取值时的值。

This will automatically add MEDIA_URL setting to the value when you fetch value.

如果您想使用 CharField ，则可以使用 SerializerMethodField ：

In case you want to use CharField you can do what you need on serializer level using SerializerMethodField:

class ProductSerializer(serializer.ModelSerializer):
    image = serializers.SerializerMethodField()

    def get_image(self, obj):
        storage_location = '/assets/img'
        img_url = f'{storage_location}/{obj.image}'
        return img_url

    class Meta:
        model = Product
        fields = '__all__'

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