Django : 動的モデル生成

  • 定義されたメタ情報で同じテーブルを複数作る
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from pathlib import Path

import pandas as pd
from django.conf import settings
from django.db import connection, models


def get_meta(name):
    """ 定義情報(CSV)ファイル """
    path = META_DIR / f"{name}.map.csv"
    return pd.read_csv(path, dtype="str").fillna("").to_dict(orient="records")


def named_model(name, code):
    meta = get_meta(name)

    class_name = f"{name}{code}"
    table_name = f"{name}_{code}"

    unique_fields = list(map(lambda i: i["column"], filter(lambda i: i["unique"] == "1", meta)))

    class Meta:
        db_table = f"oracle_{table_name}"
        constraints = [
            models.UniqueConstraint(
                fields=unique_fields,
                name=f"uniq_{table_name}",
            ),
        ]

    def _field(item):
        params = {"db_index": True} if item["db_index"] else {}
        params["blank"] = True

        return (
            item["column"],
            models.CharField(verbose_name=item["label"], max_length=100, null=True, default=None, **params),
        )

    fields = dict(map(_field, meta))

    attrs = {"__module__": "oracle.models", "Meta": Meta, **fields}
    models_class = type(class_name, (models.Model,), attrs)

    return models_class


def get_named_model(name, code):
    # モデルクラス取得
    model_class = named_model(name, code)

    if model_class._meta.db_table in connection.introspection.table_names():
        # テーブルが存在していたら終了
        return model_class

    # テーブルの作成
    with connection.schema_editor() as schema_editor:
        schema_editor.create_model(model_class)

    return model_class