我可以在Python Flask中调用带有相同文件并带有url请求的函数吗？

Question

from flask import Flask,request
import requests, json
app = Flask(__name__)
@app.route('/checking')
def test1():
    print 'in the test1 function'
    x = json.dumps({'abcd': 1234})
    requests.post('http://127.0.0.1:5123/sending', data = x)
    return 'checking done'

@app.route('/sending', methods= ['POST'])
def test2():
    print 'receiving data'
    request.data
    y = json.loads(request.data)
    print y
    return

if __name__ == '__main__':
    app.run(debug= True,host='0.0.0.0', port = 5123)

If I save this file as test.py and run it, when I call the url http://127.0.0.1:5123/checking from my browser I would expect the function test1 to send data = x to test2

Answer 1

you could do one of two things:

make a function that that returns the final data:

def test2(x):
    print('receiving data')
    print(json.loads(x))
    return y

or you could send the data in the request:

@app.route('/checking')
def test1():
    x = json.dumps({'abcd':'1234'})
    return redirect(url_for('test2', data=x))

@app.route('/sending/<data>')
def test2(data):
    print('receiving data')
    print(json.loads(data))
    return render_template('sending.html', data=data)

from flask you will need to import url_for, redirect and render_template