php表单跳转保留数据

<form id="loginform" name="loginform" action="login.php" method="post" onSubmit="return InputCheck(this)" autocomplete="off"><br />				<div id="login" class="login"><br />						<form id="loginform" name="loginform" action="login.php" method="post" onSubmit="return InputCheck(this)" autocomplete="off"><br />							<div class="inputOuter"><br />							    &nbsp&nbsp&nbsp<b>输入用户名:&nbsp&nbsp</b><input type="text" class="inputstyle" id="username" name="username" value="" tabindex="1"><br />							</div><br />							<div class="inputOuter"><br />								&nbsp&nbsp&nbsp<b>请输入密码:&nbsp&nbsp</b><input type="password" class="inputstyle password" id="password" name="password" value="" maxlength="16" tabindex="2"> <br />							</div><br />							<div class="submit"><br />								<input type="submit" tabindex="6" value="" class="btn" id="login_button"><br />							</div><br />			</form>

<?php<br />	$con = mysql_connect("localhost","root","w123456");<br />	if (!$con)<br />	  {<br />	  die('Could not connect: ' . mysql_error());<br />	  }<br /><br />		mysql_select_db("peixunrecord", $con);<br />		mysql_query("set names 'utf8'");<br /><br />	$bumen = $_POST['bumen'];<br />	$keduimokuai = $_POST['keduimokuai'];<br />	$xingming = $_POST['xingming'];<br />	$peixunshijian = $_POST['peixunshijian'];<br />	$keti = $_POST['keti'];<br />	$xinde = $_POST['xinde'];<br />	$sql="INSERT INTO record (bumen,keduimokuai,xingming,peixunshijian,keti,xinde) VALUES ('$bumen','$keduimokuai','$xingming','$peixunshijian','$keti','$xinde')";<br />	if (!mysql_query($sql,$con))<br />	  {<br />	  die('Error: ' . mysql_error());<br />	  }<br />     <br />         这块怎么能实现需求呢？谢谢了！<br /><br />	 //echo "<script>alert('提交成功!');location.href='".$_SERVER["HTTP_REFERER"]."';</script>";<br />	 //echo "提交成功 3秒后自动跳转";<br />	// echo "若没自动跳转，请点击<a href='javascript:history.go(-1)'>这里</a>返回";<br />	 //header("Refresh:5;url=http://www.baidu.com");<br />	mysql_close($con);<br />?>