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flask - Flask app-builder 如何使用文件项制作 REST API

问题描述

我正在制作一个REST api，可以像这样在flask-appbuilder中的MODEL-VIEW中上传文件。但我不知道如何调用 REST API (POST /File)。我尝试了几种不同的方法。但我不能。让我知道正确或替代方法。

[客户端代码]

file = {'file':open('test.txt', 'rb'),'description':'test'}
requests.post(url, headers=headers, files=file)

==> 失败

模型.py

class Files(Model):
    __tablename__ = "project_files"
    id = Column(Integer, primary_key=True)
    file = Column(FileColumn, nullable=False)
    description = Column(String(150))

    def download(self):
        return Markup(
            '<a href="'
            + url_for("ProjectFilesModelView.download", filename=str(self.file))
            + '">Download</a>'
        )

    def file_name(self):
        return get_file_original_name(str(self.file))

视图.py

class FileApi(ModelRestApi):
    resource_name = "File"
    datamodel = SQLAInterface(Files)
    allow_browser_login = True


appbuilder.add_api(FileApi)

标签： flaskflask-sqlalchemyflask-restfulflask-appbuilder

解决方案

FileColumn只是一个字符串字段，将文件名保存在数据库中。实际文件保存到config['UPLOAD_FOLDER']. 这由flask_appbuilder.filemanager.FileManager.

此外，ModelRestApi假设您正在发布 JSON 数据。为了上传文件，我遵循了Flask 的文档，它建议发送multipart/form-data请求。因此，需要覆盖ModelRestApi.post_headless().

这是我的解决方案，我还确保在Files删除数据库行时，文件系统中的相关文件也是如此。

from flask_appbuilder.models.sqla.interface import SQLAInterface
from flask_appbuilder.api import ModelRestApi
from flask_appbuilder.const import API_RESULT_RES_KEY
from flask_appbuilder.filemanager import FileManager

from flask import current_app, request

from marshmallow import ValidationError
from sqlalchemy.exc import IntegrityError

from app.models import Files


class FileApi(ModelRestApi):
    resource_name = "file"
    datamodel = SQLAInterface(Files)

    def post_headless(self):
        if not request.form or not request.files:
            msg = "No data"
            current_app.logger.error(msg)
            return self.response_400(message=msg)

        file_obj = request.files.getlist('file')
        if len(file_obj) != 1:
            msg = ("More than one file provided.\n"
                   "Please upload exactly one file at a time")
            current_app.logger.error(msg)
            return self.response_422(message=msg)
        else:
            file_obj = file_obj[0]

        fm = FileManager()
        uuid_filename = fm.generate_name(file_obj.filename, file_obj)
        form = request.form.to_dict(flat=True)

        # Add the unique filename provided by FileManager, which will
        # be saved to the database. The original filename can be 
        # retrieved using 
        # flask_appbuilder.filemanager.get_file_original_name()
        form['file'] = uuid_filename

        try:
            item = self.add_model_schema.load(
                    form,
                    session=self.datamodel.session)
        except ValidationError as err:
            current_app.logger.error(err)
            return self.response_422(message=err.messages)

        # Save file to filesystem
        fm.save_file(file_obj, item.file)

        try:
            self.datamodel.add(item, raise_exception=True)
            return self.response(
                201,
                **{API_RESULT_RES_KEY: self.add_model_schema.dump(
                    item, many=False),
                    "id": self.datamodel.get_pk_value(item),
                   },
                )
        except IntegrityError as e:
            # Delete file from filesystem if the db record cannot be
            # created
            fm.delete_file(item.file)
            current_app.logger.error(e)
            return self.response_422(message=str(e.orig))

    def pre_delete(self, item):
        """
        Delete file from filesystem before removing the record from the 
        database
        """
        fm = FileManager()
        current_app.logger.info(f"Deleting {item.file} from filesystem")
        fm.delete_file(item.file)

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