Flask - 从函数中获取路径URL

Question

@app.route('/this/is/the/url')
def some_exposed_func():
    return render_template("data.html")

如何从some_exposed_func句柄获取'/ this/is/the/url' ？

print 'The func URL is: %s' % get_flask_route_url(some_exposed_func)

Answer 1

找到我自己:url_for()就是答案