是否可以在Flask中发出POST请求？

Question

需要在Flask中从服务器端发出POST请求.

让我们假设我们有:

@app.route("/test", methods=["POST"])
def test():
    test = request.form["test"]
    return "TEST: %s" % test

@app.route("/index")
def index():
    # Is there something_like_this method in Flask to perform the POST request?
    return something_like_this("/test", { "test" : "My Test Data" })

我没有在Flask文档中找到任何具体内容.有人说urllib2.urlopen是问题,但我没能把Flask和urlopen.真的有可能吗？

提前致谢!

Answer 1

为了记录,这是从Python发出POST请求的通用代码:

#make a POST request
import requests
dictToSend = {'question':'what is the answer?'}
res = requests.post('http://localhost:5000/tests/endpoint', json=dictToSend)
print 'response from server:',res.text
dictFromServer = res.json()

请注意,我们使用该json=选项传入Python dict .这方便地告诉请求库做两件事:

1. 将dict序列化为JSON
2. 在HTTP标头中写入正确的MIME类型('application/json')

这是一个Flask应用程序,它将接收并响应该POST请求:

#handle a POST request
from flask import Flask, render_template, request, url_for, jsonify
app = Flask(__name__)

@app.route('/tests/endpoint', methods=['POST'])
def my_test_endpoint():
    input_json = request.get_json(force=True) 
    # force=True, above, is necessary if another developer 
    # forgot to set the MIME type to 'application/json'
    print 'data from client:', input_json
    dictToReturn = {'answer':42}
    return jsonify(dictToReturn)

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

Answer 2

是的,要发出可以使用的POST请求urllib2,请参阅文档.

但是我建议使用请求模块.

编辑:

我建议你重构你的代码来提取常用功能:

@app.route("/test", methods=["POST"])
def test():
    return _test(request.form["test"])

@app.route("/index")
def index():
    return _test("My Test Data")

def _test(argument):
    return "TEST: %s" % argument