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Python 将值从userform传递到Flask_Python_Python 2.7_Flask - Fatal编程技术网

Python 将值从userform传递到Flask

Python 将值从userform传递到Flask,python,python-2.7,flask,Python,Python 2.7,Flask,当我在这个简单的例子中单击submit按钮时,我在我的页面上得到了一个notfound错误,控制台中没有任何内容。我错过了什么 from flask import Flask app = Flask(__name__) from requests import request @app.route('/landingpage/') def landing(): return """<form action="print_output" method="post">

当我在这个简单的例子中单击submit按钮时,我在我的页面上得到了一个
notfound
错误,控制台中没有任何内容。我错过了什么

from flask import Flask
app = Flask(__name__)
from requests import request

@app.route('/landingpage/')
def landing():
    return """<form action="print_output" method="post">
    Project file path: <input type="text" name="myform"><br>
    <input type="submit" value="Submit">
    </form>"""


@app.route('/print_output', methods=['POST'])
def print_output():
    print(request.form['myform'])
    return "check your console!"
if __name__ == '__main__':
    app.run()
从烧瓶导入烧瓶
app=烧瓶(名称)
从请求导入请求
@app.route(“/landingpage/”)
def landing():
返回“”
项目文件路径:
""" @app.route('/print_output',methods=['POST']) def打印输出(): 打印(request.form['myform']) return“检查控制台!” 如果uuuu name uuuuuu='\uuuuuuu main\uuuuuuu': app.run()
有几件事需要改变:

首先,您需要从flask导入请求,这是您将收到的请求。
请求
库用于发出请求

然后,表单操作需要指向第二条路径上的url:
/print\u output

另一件需要更改的事情是
name=“\myform”
,我删除了表单上的反斜杠

还将方法大写:
method=“POST”

从烧瓶导入烧瓶
从烧瓶进口请求
app=烧瓶(名称)
@app.route(“/landingpage/”)
def landing():
返回“”
项目文件路径:
""" @app.route('/print_output',methods=['POST']) def打印输出(): 打印(request.form['myform']) return“检查控制台!” 如果uuuu name uuuuuu='\uuuuuuu main\uuuuuuu': app.run()
操作中的错误urlparam@ioganegambaputifonguser这是代码的错误之一。。。
from flask import Flask
from flask import request

app = Flask(__name__)

@app.route('/landingpage/')
def landing():
    return """<form action="/print_output" method="POST">
    Project file path: <input type="text" name="myform"><br>
    <input type="submit" value="Submit">
    </form>"""


@app.route('/print_output', methods=['POST'])
def print_output():
    print(request.form['myform'])
    return "check your console!"

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