Button to open link set in sqldb

#!/usr/bin/kivy
import kivy
kivy.require('1.7.2')

from kivy.app import App
from kivy.lang import Builder
from kivy.uix.screenmanager import ScreenManager, Screen
from kivy.uix.gridlayout import GridLayout
from kivy.uix.button import Button
import webbrowser
import pymysql.cursors

connection = pymysql.connect(host='sql6.freemysqlhosting.net',
                             user='sqlxxxx',
                             password='yyyyyy',
                             db='yyyyyyy',
                             charset='utf8mb4',
                             cursorclass=pymysql.cursors.DictCursor)
try:
    with connection.cursor() as cursor:
        # Read a single record
        sql = "SELECT `url` FROM `link` WHERE `id`=%s"
        cursor.execute(sql, ('1'))
        result = cursor.fetchone()
        print(result)
        webpage=result['url']
        print webpage
finally:
    connection.close()

Builder.load_string('''
<MenuScreen>:
    GridLayout:
        padding: 5
        spacing: 5
        cols: 1
        padding: root.width*0.1
        Button:
            background_normal: ''
            background_color:(0.862, 0.078, 0.235, 0.9)
            text: 'PLAY'
            font_size: '20sp'
            on_press: root.val1()
''')

class MenuScreen(Screen):
    def val1(self):
        global result
        print "i am executed"
        webbrowser.open(webpage)

sm = ScreenManager()
menu = MenuScreen(name='menu')
sm.add_widget(menu)

class MainApp(App):
    def build(self):
        return sm

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