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这里是我的代码:如何从一个mysql数据库中获取数据表中某个链接的行?
<html>
<?php
DEFINE('DATABASE_USER', 'sfasdfasd');
DEFINE('DATABASE_PASSWORD', 'asdfasdfasdf');
DEFINE('DATABASE_HOST', 'sdfasdfasd');
DEFINE('DATABASE_NAME', 'dsafsdfasd');
$connect = mysql_connect(DATABASE_HOST, DATABASE_USER, DATABASE_PASSWORD, DATABASE_NAME) or
die ("Hey loser, check your server connection.");
mysql_select_db("minedb");
$query = "SELECT * FROM ideastable ORDER BY datee DESC";
$quey1="select * from ideastable";
$result=mysql_query($query) or die(mysql_error());
?>
<table border=1 style="background-color:#000000;" >
<caption><EM>Ideas List</EM></caption>
<tr>
<th>IDEAS</th>
<th>Thumbs Ups</th>
</tr>
<?php
while($row=mysql_fetch_array($result)){
echo "</td><td>";
echo $row['idea'];
echo "</td><td>";
echo $row['thumbsup'];
$i = $row['id'];
echo $i;
echo '<a href="#" onClick="doSomething()">Thumbs Up!</a>';
echo "</td></tr>";
}
echo "</table>";
?>
<SCRIPT type="text/javascript" src="jquery.min.js"></SCRIPT>
<SCRIPT type="text/javascript">
function doSomething() {
var myVar = "<?php echo $i; ?>";
alert(myVar);
$.load('uts.php?i=myVar');
return false;
}
</SCRIPT>
</html>
我的问题是我怎么就能够让这个当我点击了大拇指它承认该行的链接是?我正在制作一个网站,您可以评估数据库中的一些对象,这是它的启动。