如何显示“请稍候...”在网页上单击使用Python和HTML的按钮
如何显示“请稍候...”在网页上单击使用Python和HTML的按钮
提问于 2012-11-13 06:03:15
我需要显示文本“请稍候...”当我点击on按钮，直到guvcview打开，然后显示"guvcview is .The running“，而不是关闭下面的guvcview?.In代码时，我试图显示”请稍候...“。但是我不能。有些人说这需要重新加载页面。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)回答 1
Stack Overflow用户
回答已采纳
发布于 2012-11-14 01:03:54
就像其他人建议的那样，你将需要客户端脚本来让它工作。但是，如果实现了ajax解决方案，则不需要重定向到另一个页面。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)希望这能有所帮助！
安德鲁
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/13352332
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