实现一个基于Ajax的调查程序- AJAX技术

当前位置: 首页 > 图文教程 > 网络编程 > AJAX技术 > 实现一个基于Ajax的调查程序

技术文章搜索

关键字

AJAX技术中的实现一个基于Ajax的调查程序

出处:互联网 整理: 软晨网（RuanChen.com） 发布: 2009-11-03 浏览: 99 ::

收藏到网摘: n/a

AJAX结合CSS实现的网页Tab选项卡效果 ASP.NET+AJAX解决网页打开等待问题

<html>
<head>
<title>投票</title>
<META http-equiv=Content-Language content=zh-cn>
<META http-equiv=Content-Type content="text/html; charset=gb2312">
<style type="text/CSS">

</style>
<?php
include_once("server1.server.php"); #servidor para XAJAX
$xajax->printjavascript();
?>
</head>
<body>
<script language=javascript>
    function back() {
      document.getElementById(’poll’).style.display = ’block’;
      document.getElementById(’pollresult’).style.display = ’none’;
      document.getElementById(’pollresult’).innerHTML = ’’;
    }
</script>
<div id=pollresult class=pollresult>
</div>

<?php

global $db;
$poll = $db->getRow("select * from TBL_POLL order by poll_id desc limit 1");
$poll_id = $poll["poll_id"];
$pollitems = $db->getAll("select * from TBL_POLLITEM where poll_id=$poll_id");
?>
<div id=poll class=poll>
<form id="pollForm" action="javascript:void(null);" onsubmit="onSubmit();">
   <?php echo $poll["title"]; ?><br>
   <?php for ($i = 0, $count = count($pollitems); $i < $count; $i++) { ?>
  <input type="radio" style="background-color : #CCCCCC;" name="pollitem" value="<?php echo $pollitems[$i][’pollitem_id’] ?>"><?php echo $pollitems[$i][’content’] ?><br>
  <?php } ?>
  <input type="hidden" name="poll_id" value="<?php echo $poll_id; ?>">
  <input type="submit" value="enter">
</form>
<script language=javascript>
    function onSubmit() {
      xajax_poll(xajax.getFormValues("pollForm"));
      document.getElementById(’poll’).style.display = ’none’;
      document.getElementById(’pollresult’).style.display = ’block’;
    }
</script>
</div>
</body>
</html>
服务器端
function poll($formData){
  global $db;
  $tmp="";
  $objResponse = new xajaxResponse();

  $poll_id = $formData[’poll_id’];

  $pollitem_id = $formData[’pollitem’];

  if($pollitem_id > 0 && $poll_id > 0) {
   $db->query("update ".TBL_POLLITEM." set count=count+1 where pollitem_id = $pollitem_id");
  }

  $poll = $db->getRow("select * from TBL_POLL where poll_id = $poll_id");
  $pollitems = $db->getAll("select * from TBL_POLLITEM where poll_id=$poll_id");


  $tmp&nbs

AJAX结合CSS实现的网页Tab选项卡效果 ASP.NET+AJAX解决网页打开等待问题

AJAX技术 中的 实现一个基于Ajax的调查程序

评论 (0) All

AJAX技术中的实现一个基于Ajax的调查程序