使用标识与POST在数据库中创建新任务

Question

我想根据我将放在URL中的todo_ID在我的数据库中创建一个新任务。我可以完全在数据库中发布一个新任务，但我希望服务器选择todo_ID，因此如果某个人在url中输入/ 1，它将创建一个todo_ID = 1的新任务，并根据哪些内容填写UserID和详细信息用户将在问题是我不知道该怎么办呢基础上，todo_ID所以有人告诉我如何：d

from flask import Flask, jsonify,json, request, abort 
from flask_sqlalchemy import SQLAlchemy 

app = Flask(__name__) 
app.config.from_pyfile('Config.py') 
db = SQLAlchemy(app) 


class User(db.Model, JsonModel): #Class which is a model for the User table in the database 
    User_ID = db.Column(db.Integer, primary_key = True) 
    FirstName = db.Column(db.String(20)) 
    LastName = db.Column(db.String(20)) 

    def __init__(self,User_ID,FirstName, LastName): 
     self.User_ID = User_ID 
     self.FirstName = FirstName 
     self.LastName = LastName 

class Todo(db.Model, JsonModel): #Class which is a model for the Todo table in the database 
    todo_ID = db.Column(db.Integer, primary_key = True) 
    UserID = db.Column(db.Integer, db.ForeignKey("user.User_ID")) 
    details = db.Column(db.String(30)) 

    def __init__(self, UserID, details): 
     self.UserID = UserID 
     self.details = details 


    @app.route('/<int:todo_ID>', methods = ['POST']) #Uses POST method with same URL as GET method to add new information to Todo table. 
def create_dev(todo_ID): 
    id = Todo.query.create(todo_ID) 
    dev = request.json["UserID"],request.json["details"] 
    db.session.add(id,dev) 
    db.session.commit() 
    return "New task has been created.", 201 


@app.before_first_request #Creates everything before the first request. 
def startup(): 
    db.create_all() 

if __name__ == '__main__': 
    app.run() 

Answer 1

尝试

dev.UserID = request.POST["UserID"] 

对于request.json工作，设置内容键入到application/json。从Flask documentation

如果mimetype是application/json，它将包含解析的JSON数据。否则，这将是无。

有关详细信息：https://stackoverflow.com/a/20001283/3930114

编辑1：解决方案基于在评论

@app.route('/<int:todo_ID>', methods = ['POST']) #Uses POST method with same URL as GET method to add new information to Todo table. 
# Setting todo_ID default to None 
def create_dev(todo_ID=None): 
    # This if will allow you to not pass id via url and handle in backend. Frankly, first integer field will automatically become autoincrement field in Flask, so I am not even sure why you want to send todo_ID via url but I dont have enough information about your app :) 
    if todo_ID is None: 
     # write logic to create a todo_ID 
     pass 
    # Instance of model task 
    dev.UserID = request.json["UserID"] 
    dev.details = request.json["details"] 
    t = Task(todo_ID, dev.UserID, dev.details) 
    db.session.add(t) 
    db.session.commit() 
    return "New task has been created with todo_ID: " + todo_ID, 201 

编辑2的讨论：基于在评论其他信息

@app.route('/', methods = ['POST']) #Uses POST method with same URL as GET method to add new information to Todo table. 
# Setting todo_ID default to None 
def create_dev(): 
    # Instance of model task 
    t = Todo(UserID=request.json["UserID"], details=request.json["details"]) 
    db.session.add(t) 
    db.session.commit() 
    return "New task has been created with todo_ID: " + todo_ID, 201 

PS解决方案：我没有测试过这个代码。

参考文献：

• http://flask-sqlalchemy.pocoo.org/2.1/queries/
• Flask Optional URL parameters