0
<code>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <title>jQuery.pager.js Test</title>
    <link href="Pager.css" rel="stylesheet" type="text/css" />
    <script src="jquery-1.2.6.min.js" type="text/javascript"></script>
    <script src="jquery.pager.js" type="text/javascript"></script>
    <script type="text/javascript" language="javascript">

        $(document).ready(function() {
            $("#pager").pager({ pagenumber: 1, pagecount: 15, buttonClickCallback: PageClick });
        });

        PageClick = function(pageclickednumber) {
            $("#pager").pager({ pagenumber: pageclickednumber, pagecount: 15, buttonClickCallback: PageClick });
            $("#result").html("Clicked Page " + pageclickednumber);
        }

    </script>

</head>
<body>

$query = "select name from student";
$result = mysql_query(Query);
while($row=mysql_fetch_array($result)){
$student_name = $row['name'];
?>
<h1 id="result"><?php echo $student_name; ?></h1>
<? }
?>

<div id="pager" />
</body>
</html>
</code>

对于我上面的代码没有得到学生姓名,如果我删除寻呼脚本,学生姓名会出现,我可以知道,为什么,我在哪里出错了..

我必须将 somthing 传递给 html() ,但我不确定..

4

1 回答 1

0
while($row=mysql_fetch_array($result)){ $student_name = $row['name']; }

应该

$student_name = NULL;
while($row=mysql_fetch_array($result)){ $student_name .= $row['name']; }
于 2010-01-18T13:52:54.620 回答