Many to One association in Flask / SQLAlchemy-CodePudding

I have two classes one is Dossier and the other is File I have an association Many to One (Many files can belong to one dossier) I have a problem when I try to upload a file I need to have the id of the dossier but I just couldn't get it right okay let me explain more this is my code: These are my classes:

from io import BytesIO
from xml.dom.minidom import Identified
from flask import Flask, abort, render_template, request, send_file
from flask_sqlalchemy import SQLAlchemy 
import os
from flask_restful import Api, Resource, reqparse 
from flask_cors import CORS, cross_origin
 
app = Flask(__name__)
CORS(app)

app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///db.sqlite3'
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False 
db = SQLAlchemy(app)
api = Api(app)

class Fichier(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    filename = db.Column(db.String(50))
    data = db.Column(db.LargeBinary)
    type=db.Column(db.String(50))
    size=db.Column(db.Integer)
    idDossier = db.Column(db.Integer, db.ForeignKey('dossier.idD'))
    def __init__(self, filename, data, type, size, idDossier):
        self.filename = filename
        self.data = data
        self.type = type
        self.size = size
        self.idDossier = idDossier
class Dossier(db.Model):
    idD=db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(50))
    idChild= db.Column(db.Integer)
    idFichier = db.relationship('Fichier',  backref='dossier')
    def __init__(self, idD , name , idChild ): 
        self.idD = idD
        self.name = name
        self.idChild = idChild
    
    def json(self): 
        return {"id":self.id , "filename":self.filename , "type":self.type , "size":self.size}

    def toJson(self):
        return {"name":self.name , "idD":self.idD , "idChild":self.idChild}

This is my upload function :

    @app.route('/upload/<id>', methods=['POST'])
     def index(id):
        file = request.files['file']
        file.data = request.files['file'].read()
        file.size=len(file.data)
        type = os.path.splitext(file.filename)
        file.type = type[1]
        file.size=len(file.data)
        file.idDossier = Dossier.query.filter_by(idD=id).first()
        """file.idDossier= Dossier.query.get(Dossier.idD)"""
        upload = Fichier(filename=file.filename, data=file.data , 
        type=file.type , size=file.size , idDossier=file.idDossier)
        db.session.add(upload) 
        db.session.commit()
        return f'Uploaded: {file.filename}'

when I run my code , I get this error in console :

I'm lost if anyone knows how to fix this that would be great. Thank you!

CodePudding user response：

You need to pass the id of the idDossier, but that's confusing.

upload = Fichier(
    filename=file.filename,
    data=file.data,
    type=file.type,
    size=file.size,
    idDossier=file.idDossier.idD,
)

According to sqlalchemy documentation, this is the pattern you should be following. Keep it simple, clear and formatted, you will read it easier for sure.

class Fichier(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    ...
    dossier_id = db.Column(db.Integer, db.ForeignKey("dossier.id"))


class Dossier(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    ...
    fichier = db.relationship("Fichier", backref="dossier")

Side note: maybe try to avoid backref for a more explicit code ?