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我知道如何返回一個位置標題,其URL爲說「todo/1」,我必須在我的標題位置輸入一個,這樣我的代碼將如下所示POST方法。不過,我不知道如何根據todo_ID返回它,所以我不必手動輸入它。因此,例如它看起來像在燒瓶中獲取位置標題返回一個標識
response.headers['location'] = '/todo/todo_ID'
但是,這將返回單詞todo_ID。無論如何,我可以返回已在網址中創建的實際todo_ID?
我看着這個問題,但不知道如果答案會幫助我。 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
你檢查你有'todo'? – furas
不確定你的意思。如果你在討論URL中的/ todo,那麼它只是網址的一部分。 – Muba
我在'todo = Todo(..)'中要求'todo'' – furas