实现一个基于Ajax的调查程序

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Ajax PHP JavaScript CSS HTML

<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>

</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

数据库的表如下

CREATE TABLE TBL_POLL (
poll_id int(11) unsigned NOT NULL default '0',
title varchar(100) NOT NULL default '',
created_date bigint(20) unsigned NOT NULL default '0',
user_id int(11) unsigned NOT NULL default '0',
PRIMARY KEY (poll_id)
) TYPE=MyISAM;

CREATE TABLE TBL_POLLITEM (
pollitem_id int(11) unsigned NOT NULL default '0',
poll_id int(11) unsigned NOT NULL default '0',
content varchar(100) NOT NULL default '',
count int(11) unsigned NOT NULL default '0',
PRIMARY KEY (pollitem_id)
) TYPE=MyISAM;

这个例子中，调查的选项只在页面装载时读出，投票后在原地显示最新的投票信息。不需要弹出窗口

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