在烧瓶中获取位置标题返回一个标识

Question

我知道如何返回一个位置标题，其URL为说“todo/1”，我必须在我的标题位置输入一个，这样我的代码将如下所示POST方法。不过，我不知道如何根据todo_ID返回它，所以我不必手动输入它。因此，例如它看起来像

response.headers['location'] = '/todo/todo_ID' 

但是，这将返回单词todo_ID。无论如何，我可以返回已在网址中创建的实际todo_ID？

我看着这个问题，但不知道如果答案会帮助我。 How to return a relative URI Location header with Flask?

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

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

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('/todo', methods = ['POST']) #Uses POST method with same URL as GET method to add new information to Todo table. 
def create_todo(): 
    if not request.json: 
     abort(400) 
    response= jsonify() 
    todo = Todo(UserID = request.json["UserID"],details = request.json["details"]) 
    db.session.add(todo) 
    db.session.commit() 
    response.status_code = 201 
    response.headers['location'] = '/todo/1' 
    return response 

Answer 1

您需要的几个步骤。

1. db.session.commit（）将删除对象，使用冲洗第一
2. response.headers [ '位置'] =“/todo/{}'.format(todo.todo_ID）
3. 和然后db.session.commit（）

这里是整个代码：

@app.route('/todo', methods = ['POST']) 
def create_todo(): 
    if not request.json: 
     abort(400) 
    response= jsonify() 
    todo = Todo(UserID = request.json["UserID"],details = request.json["details"]) 
    db.session.add(todo) 
    #db.session.commit() 
    db.session.flush() # will get id from database 
    response.status_code = 201 
    response.headers['location'] = '/todo/{}'.format(todo.todo_ID) 
    db.session.commit() # write to database 
    return response