Flask app-builder how to make REST API with file items

Question

I'm making a REST api that files can be uploaded based in MODEL-VIEW in flask-appbuilder like this. But I don't know how to call REST API (POST /File). I tried several different ways. but I couldn't. Let me know the correct or the alternative ways.

[client code]

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

==> Failed

model.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))

view.py

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


appbuilder.add_api(FileApi)

score 1 · Answer 1 · answered May 05 '21 at 14:46

FileColumn is only a string field that saves the file name in the database. The actual file is saved to config['UPLOAD_FOLDER']. This is taken care of by flask_appbuilder.filemanager.FileManager.

Furthermore, ModelRestApi assumes that you are POSTing JSON data. In order to upload files, I followed Flask's documentation, which suggests to send a multipart/form-data request. Because of this, one needs to override ModelRestApi.post_headless().

This is my solution, where I also make sure that when a Files database row is deleted, so is the relative file from the filesystem.

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)

Thanks! It is an excellent Approach to me. – Hyoungki Kim May 05 '21 at 23:59 — Hyoungki Kim, May 05 '21 at 23:59

score 0 · Answer 2 · answered Apr 07 '21 at 13:53

You can use this.

from app.models import Project, ProjectFiles

class DataFilesModelView(ModelView):
datamodel = SQLAInterface(ProjectFiles)

label_columns = {"file_name": "File Name", "download": "Download"}
add_columns = ["file", "description", "project"]
edit_columns = ["file", "description", "project"]
list_columns = ["file_name", "download"]
show_columns = ["file_name", "download"]

Last add the view to the menu.

appbuilder.add_view(DataFilesModelView,"File View")

Thanks for helpful answer. but I need REST /POST API to upload a file. — Hyoungki Kim, Apr 12 '21 at 23:42

Flask app-builder how to make REST API with file items

2 Answers2