我们说我有一个创建$res=mysqli_query($dbh, "SELECT column_b FROM table_b");
while ($r=mysqli_fetch_assoc($res)) {
$ins="INSERT INTO table_a(column_a) VALUES ("
. multi($r['column_b']) , ");";
if (!mysqli_query($dbh, $ins)) {
die mysqli_error($dbh);
}
}
的端点需要一个Player对象：
Team我有一个ModelViewSet如下：
class Team(Model):
name = CharField()
class Player(Model):
name = CharField()
team = ForeignKey(Team)
class PlayerViewSet(ModelViewSet):
model = Player
serializer = PlayerSerializer
当我看到我的Django Rest Swagger时，＆＃34;示例值＆＃34;对于数据，我需要class TeamSerializer(ModelSerializer):
class Meta:
model = Team
fields = "__all__"
class PlayerSerializer(ModelSerializer):
class Meta:
model = Player
fields = "__all__"
来创建一个玩家：
POST我希望扩展{
"name": "string",
"team": {}
}
，以表明我需要提供{}对象来创建{"name": "string"}对象。我尝试过添加
Team
到team = TeamSerializer(help_text="{'name': 'string'}")，但只将数据添加到Django Rest Swagger上的PlayerSerializer标签中。有没有办法自动将此文档添加到Model标签，Example显示所需的Model数据？