Question

所以我有以下app.py文件，我试图处理一个帖子请求，它基本上会读取名字，姓氏，用户名和密码，然后将它存储在mongodb中，但这是为了以后。截至目前，我无法收到发布/获取请求。

当我尝试使用postman将数据提交到localhost：5000 / register时，即使是我尝试运行的小片段也会给我一个错误的请求错误。

任何帮助将不胜感激。

from flask import *
#from pymongo import MongoClient
#import json
app = Flask(__name__)

# main interface
@app.route("/")
def main():
    return render_template('index.html')

# Register Interface
@app.route('/register/', methods = ['GET'])
def register():
    #collection = db['userdb']
    firstname = request.form['firstname']
    lastname = request.form['lastname']
    username = request.form['username']
    password = request.form['password']
    #postData = { 'firstname': firstname, 'lastname': lastname,'username':username,'password':password}
    #json = json.dumps(postData)
    #try:
        #status = db.userdb.insert_one(postData).insert_id
    #except:
        #status = 'This user is already registerd'
    return 'This data works'

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

Answer 1

你需要添加＆＃34; POST＆＃39;路径装饰器的方法，但还需要一个步骤。使用if语句使视图忽略request.form赋值：

from flask import *
#from pymongo import MongoClient
#import json
app = Flask(__name__)

# main interface
@app.route("/")
def main():
    return 'go to <a href="/register/">register</a>'

# Register Interface
@app.route('/register/', methods = ['GET','POST'])
def register():
    #collection = db['userdb']
    if request.method == 'POST':
        firstname = request.form['firstname']
        lastname = request.form['lastname']
        username = request.form['username']
        password = request.form['password']
        #postData = { 'firstname': firstname, 'lastname': lastname,'username':username,'password':password}


    return 'This data works'

if __name__ == "__main__":
    app.run(debug=True)

这是使用GET和POST的视图中的常见模式。另请注意，我已将debug = True添加为app.run的参数。这将使调试代码变得更加容易。

Answer 2

@app.route('/register/', methods = ['GET'])

上面的装饰器意味着处理程序只会映射到GET请求。

@app.route('/register/', methods = ['GET', 'POST'])

尝试一下，你应该好好去。

烧瓶和POST / GET的问题

2 个答案: