Question

我需要显示文字“Please wait .....”当我点击ON按钮直到guvcview打开，然后显示“guvcview正在运行”。当我关闭guvcview时同样的事情？下面的python代码，我试图显示“请等待.....”但我不能。有人说需要重新加载页面。这只是一个例子，在我的代码中，我对页面登录和注销进行了身份验证。我需要最简单的方法，谢谢。

import cherrypy
import os.path
import struct
import time
import subprocess
import commands

class Server(object):
    led_on=1 
    led_off=1 
    def index(self,  on='', off=''):
        html = """
         <html>
           <body>
             <br>
             <p>{htmlText}
             <p>
             <a href="?on=1"><img src="images/on.png"></a>
             <a href="?off=1"><img src="images/off.png"></a>
           </body>
          </html>
                """
        myText = ''
        if on:
            self.led_on = int(on)             
            myText = "Please wait ....."
            html.format(htmlText=myText)
            subprocess.call(['guvcview &'], shell=True)
            time.sleep(2)
            output = commands.getoutput('ps -A')
            if 'guvcview' in output:
                myText = "guvcview is running"

        if off:
            self.led_off = int(off)             
            myText = "Please wait ....."
            html.format(htmlText=myText)
            subprocess.call(['sudo pkill guvcview'], shell=True)
            time.sleep(2)
            output = commands.getoutput('ps -A')
            if 'guvcview' in output:
                myText = "Please wait ....."
            else:
                myText = "guvcview closed"

        return html.format(htmlText=myText)
    index.exposed = True
conf = {
        'global' : { 
            'server.socket_host': '0.0.0.0', 
            'server.socket_port': 8085 
        },

        '/images': {
            'tools.staticdir.on': True,
            'tools.staticdir.dir': os.path.abspath('images')
        }
    }
cherrypy.quickstart(Server(), config=conf)

Answer 1

像其他人一样建议你需要客户端脚本来实现这一点。但是，如果实现ajax解决方案，则无需重定向到其他页面。试一试，看看这是不是你想要的。

import cherrypy
import os.path
import struct
import time
import subprocess
import commands

class Server(object):
    led_on=1 
    led_off=1 
    def index(self):
        html = """
         <html>
           <body>
           <script language="javascript" type="text/javascript">
           function Activate(CurrentState)
           {
               // code for IE7+, Firefox, Chrome, Opera, Safari
               if(window.XMLHttpRequest)
                   xmlhttp=new XMLHttpRequest();
               else// code for IE6, IE5
                   xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");

               xmlhttp.onreadystatechange=function()
               {
                   if (xmlhttp.readyState==4 && xmlhttp.status==200)
                   {
                       document.getElementById("UserMessage").innerHTML = xmlhttp.responseText;
                   }
               }

               xmlhttp.open("GET","/Process?on=" + CurrentState, true);
               xmlhttp.send();
           }
           </script>
             <br>
             <p id="UserMessage"><p>
             <a onclick="document.getElementById('UserMessage').innerHTML = 'Please wait .....';Activate('1');"><img src="images/on.png"></a>
             <a onclick="document.getElementById('UserMessage').innerHTML = 'Please wait .....';Activate('0');"><img src="images/off.png"></a>
           </body>
          </html>
                """
        return html
    index.exposed = True

    def Process(self,  on='0'):
        if on == '1':
            self.led_on = int(on)             
            subprocess.call(['guvcview &'], shell=True)
            time.sleep(2)
            output = commands.getoutput('ps -A')
            if 'guvcview' in output:
                return "guvcview is running"

        if on == '0':
            self.led_off = int(on)
            subprocess.call(['sudo pkill guvcview'], shell=True)
            time.sleep(2)
            output = commands.getoutput('ps -A')
            if 'guvcview' in output:
                return "Please wait ....."
            else:
                return "guvcview closed"

        return "Please wait ....."
    Process.exposed = True

conf = {
        'global' : { 
            'server.socket_host': '0.0.0.0', 
            'server.socket_port': 8085 
        },

        '/images': {
            'tools.staticdir.on': True,
            'tools.staticdir.dir': os.path.abspath('images')
        }
    }
cherrypy.quickstart(Server(), config=conf)

希望这有帮助！

安德鲁

如何在使用Python和HTML单击按钮时在网页上显示“请稍候....”

1 个答案: