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這裏是我的代碼：如何從一個mysql數據庫中獲取數據表中某個鏈接的行？
<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>
我的問題是我怎麼就能夠讓這個當我點擊了大拇指它承認該行的鏈接是？我正在製作一個網站，您可以評估數據庫中的一些對象，這是它的啓動。