代码之家 › 专栏 › 技术社区 › Kevin Müller

从Flask中的url获取变量以在函数外部使用

flask python

-1

Kevin Müller · 技术社区 · 7 年前

我是个编程初学者,希望能在这里得到一些帮助。

我的问题是: 我有这个密码:

import os, json, subprocess, sys
import CallExternalServiceConfig as c
from flask import Flask, jsonify

DATAS = "datas"
SOURCE_FOLDER = "source_folder"
NAME_SCRIPT = "name_script"
PARAMETERS = "parameters"

with open("CallScript.json", "r") as read_file:
data = json.load(read_file)

for trigger in data[DATAS]:
adress=trigger[SOURCE_FOLDER]
namescript=trigger[NAME_SCRIPT]
parameter=trigger[PARAMETERS]

app= Flask(__name__)

@app.route('/Test/<string:jsonname>/')

def action():
    #try:
    subprocess.call([adress+namescript, parameter.split(",")], shell=True)
    print(parameter.split(","))
    return jsonify(adress+namescript)
if __name__ == "__main__":
    app.run(host=c.get(c.SERVER_HOST, '127.0.0.1'), port=int(c.get(c.SERVER_PORT, 
    '5000')));

我想把 jsonname 字符串,并使用它替换 "CallScript.json" 所以它是动态的。但我不知道如何访问 def action() 功能。

我在用烧瓶

2 回复 | 直到 7 年前

1

0

Srikanth Chekuri 7 年前

import os, json, subprocess, sys
import CallExternalServiceConfig as c
from flask import Flask, jsonify

DATAS = "datas"
SOURCE_FOLDER = "source_folder"
NAME_SCRIPT = "name_script"
PARAMETERS = "parameters"


app= Flask(__name__)

@app.route('/Test/<string:jsonname>/')

def action():
    json_name = request.args.get('jsonname', default = '', type = str)
    with open(json_name + ".json", "r") as read_file:
        data = json.load(read_file)
    for trigger in data[DATAS]:
        adress=trigger[SOURCE_FOLDER]
        namescript=trigger[NAME_SCRIPT]
        parameter=trigger[PARAMETERS]

    subprocess.call([adress+namescript, parameter.split(",")], shell=True)
    print(parameter.split(","))
    return jsonify(adress+namescript)
if __name__ == "__main__":
    app.run(host=c.get(c.SERVER_HOST, '127.0.0.1'), port=int(c.get(c.SERVER_PORT, 
    '5000')));

2

0

Windmill 7 年前

首先,我们初始化您希望能够在函数外部访问的变量。然后,在调用URL时,我们将json文件名传递给一个函数,该函数将这些参数重置为该json文件中的值。

adress = ""
namescript = ""
parameter = ""

def read_json(filename):
    with open(filename, "r") as read_file:
        data = json.load(read_file)
    for trigger in data[DATAS]:
        adress=trigger[SOURCE_FOLDER]
        namescript=trigger[NAME_SCRIPT]
        parameter=trigger[PARAMETERS]

@app.route('/Test/<string:jsonname>/')
def action():
    json_name = request.args.get('jsonname', default = '', type = str) + ".json"
    read_json(json_name)

当你的路线被调用时, json_name 现在将被设置为URL中的任何参数。