Flask url_for 在从表单中提取返回值时传递 arguments 错误

Question

所以我实现了这段代码：

from flask import Flask, render_template,redirect,url_for
from flask.globals import request
from task2_code import predict_num_new_cases as p


app=Flask(__name__)


@app.route('/')
def index():
    return render_template('index.html')

@app.route('/Task1')
def Task1():
    return render_template('Task 1.html')

@app.route('/Task2', methods=['GET','POST'])
def Task2():
    render_template('Task 2.html')
    if request.method=='POST' or request.method=='GET':
        nvax=request.form.get('nvax')
        po=request.form.get('pop')
        co=request.form.get('c')
        return redirect(url_for('display',novax=nvax,pop=po,country=co))
    else:
        return render_template('Task 2.html')

@app.route('/pdfview1')
def pdffor1():
    return render_template

@app.route('/Task 2/display')
def display(novax,pop,country):
    predicted=p(novax,pop,country)
    return render_template('display.html',predicted)


@app.route('/bibliography')
def bibliography():
    render_template('bibliography.html')


'''def functionultimate(no_cases,population):
    #calculate stuff
    return #a string value that can be concatenated into task 2'''

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

但它会引发以下错误：

TypeError: display() missing 3 required positional arguments: 'novax', 'pop', and 'country'

My goal is to pass in arguments into the display function such that it invokes another function from task2_code and displays the result in display.html How else do I pass in arguments for the display function?

Answer 1

参数是通过请求传入的，因此您必须从请求中提取它们。 尝试这样的事情。

@app.route('/Task 2/display')
def display():
    novax = request.args.get('novax')
    pop = request.args.get('pop')
    country = request.args.get('country')
    predicted=p(novax,pop,country)
    return render_template('display.html',predicted)

如何将 arguments 传递到 Flask 的 redirect(url_for()) 中？