php 怎样做弹出窗口，弹出窗口中显示数据库表查询记录，哪位高手有代码参考;
<script LANGUAGE="JavaScript">
<!--
function openwin() {
window.open ("list_conn_dj.php", "newwindow", "height=200, width=400, toolbar =no, menubar=no, scrollbars=yes, resizable=no, location=no, status=no, top="+(window.screen.height-450)/2+", left="+(window.screen.width-450)/2+" ")
// this.Response.Write("<script language=javascript> window.open ("list_conn_dj.php", "newwindow", "height=200, width=400, toolbar =no, menubar=no, scrollbars=yes, resizable=no, location=no, status=no, top="+(window.screen.height-450)/2+", left="+(window.screen.width-450)/2+" ") </script>")
}
//-->
</script>
最新回答
仙女不仙
2024-06-26 01:24:10
<head>
<script ...>
function send_request()
{
//ajax发送查找数据库请求
//设置onready函数等待数据接收完成
//当数据接收完成,将<div id="data_recv"></div>之间的innerHtml设置为接受到的数据
//并显示该div(初始化时设置为隐藏)
}
</script>
</head>
<body>
//搜索部分
<div id="data_recv">
</div>
</body>
</html>
......php文件
根据发送来的请求搜索数据库,并生成相应的 html,然后echo 之
菊花般的笑容
2024-06-26 12:05:52
$query = mysql_query($sql);
$count = mysql_result($query);
echo "<script>alert(".$count.")</script>";
深情是罪
2024-06-26 10:32:46
元亦夏
2024-06-26 03:06:20
热门标签
