爱钓猫的饕餮 爱钓猫的饕餮
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【求助】苦逼新手伤不起。页面无法获得表单数据,也无法跳转... RT 不多说 上代码 有点多。。我会分楼发。求帮忙啊啊 啊。。 <? //初始化session session_cache_expire(30); session_start(); error_reporting(E_ALL & ~ E_NOTICE); // 如果没有登录,退出 if(!isset($_SESSION['user'])) { echo "<p align=center>"; echo "<font color=#FF0000 size=5><strong><big>"; echo "您还没有登录,请<a href='login.php'>登录</a>!"; echo "</big></strong></font></p>"; exit(); } include ('head.php'); require ('dbconnect.php'); ?> <script language="JavaScript"> function select_change(){ form1.submit(); } </script> <html> <body> <?php // 如果是提交前 if (($reg=="")&&($mod=="")) { ?> <form name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF'] ?>" > <table width="60%" border="0" cellspacing="1" cellpadding="3" align="center" bordercolor="#8695AC"> <tr> <th colspan="2">厂 商 入 库 管 理</th> </tr> <tr align="center"> <td colspan="2">&nbsp;</td> </tr> <tr align="center"> <td colspan="2"> <select name="show" onChange="select_change()"> <option value="" selected>查看已入库的厂商信息</option> <?php $sql="SELECT * FROM company"; $result = mysql_query($sql,$conn) or die("数据库操作错误"); while ($userrow=mysql_fetch_Array($result)){ // 显示该厂商信息 if ($show==$userrow[id]){ $id=$userrow[id]; $name=$userrow[name]; $address=$userrow[address]; $tel=$userrow[tel]; $linkman=$userrow[linkman]; $product=$userrow[product]; echo "<option value=$userrow[id] selected>$userrow[id]:$userrow[name]</option>"; } else { echo "<option value=$userrow[id]>$userrow[id]:$userrow[name]</option>"; } } ?>
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