Flask应用程序生成器如何使用文件项创建REST API

Question

我正在制作一个REST api，可以像这样在flask appbuilder中基于MODEL-VIEW上传文件。但是我不知道如何调用REST API(POST/File(。我尝试了几种不同的方法。但我做不到。让我知道正确的或替代的方法。

【客户代码】

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

===>失败

型号.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)

Answer 1

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

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

这是我的解决方案，我还确保当Files数据库行被删除，文件系统中的相关文件也被删除。

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)

Answer 2

您可以使用它。

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"]

最后将视图添加到菜单中。

appbuilder.add_view(DataFilesModelView,"File View")