PHP 菜鸟写的代码
菜鸟写的留言板登陆代码各位大神帮我看看,为什么要点击登陆两次才能登陆,为什么要点击两次退出才能退出?问题出在哪里，大神帮我看看，先谢谢你了。代码如下：<?phpinclude("Sql.php");if ($_GET[out]){ setcookie("cookie","out"); echo "<script language=\"javascript\">localhost.href='login.php';</script>";}if ($_POST[id]=='xxx') { $pw=md5($_POST[pw]); if($pw=='e1bfd762321e409cee4ac0b6e841963c') { setcookie("cookie","ok"); echo "<script language=\"javascript\">localhost.href='login.php';</script>"; } }include("Head.php");if ($_COOKIE['cookie']!='ok'){?><SCRIPT language=javascript>function Checklogin() { if (myform.id.value=="") { alert ("请填写登陆名"); myform.id.focus(); return false; } if (myform.pw.value=="") { alert ("密码不能为空"); myform.pw.focus(); return false; } }</SCRIPT> <form action="" method="post" name="myform" onsubmit="return Checklogin();"> ID:<input type="text" name="id" ><br> PW:<input type="password" name="pw"> <input type="submit" name="submit" value="登陆" > </form><?}else{?> <a href='?out=login'>退出</a><?}?> 回复讨论(解决方案)
不知错哪了，望大神指点迷经啊
在纠正了“变量未定义”的警告后，未发现你说的现象
但不能排除是 sql.php 或 head.php 中的问题
贴出你include的文件看下、看是否异样。
贴出你include的文件看下、看是否异样
Sql.php文件代码:
<?php/*CREATE TABLE `message` ( `id` tinyint(1) NOT NULL auto_increment, `user` varchar(25) NOT NULL, `title` varchar(50) NOT NULL, `content` tinytext NOT NULL, `lastdate` date NOT NULL, PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=gbk AUTO_INCREMENT=1 ;*/$conn=mysql_connect("localhost","root","") or die ("数据库错误");mysql_select_db("bbs",$conn);mysql_query("set names 'GBK'"); //中文编码function htmtocode($content) { $content = str_replace("\n","<br>",str_replace(" "," ",$content)); return $content; }//把输入的空格跟换行用str_replace("\n","<br>",str_replace(" "," ",$content))替换成PHP能接受的格式,实现输入什么格式就显示什么格式.?> Head.php文件代码:
<head><meta http-equiv="Content-Type" content="text/html";charset="gb2312"><title></title><link href="images/css.css" rel="stylesheet" type="text/css"><b><a href="Textbook.php">添加留言</a> | <a href="List.php">浏览留言</a> | <a href="login.php">登陆</a> </b><hr size=1>
Textbook.php文件代码：
<?phpinclude("Sql.php");//if ($_POST['submit'])if (isset($_POST['submit'])) //注一定用(isset($_POST['submit']))因为它是判断 submit的数据提交没有.{ $sql="insert into message(id,user,title,content,lastdate)values('','$_POST[user]','$_POST[title]','$_POST[content]',now())"; mysql_query($sql); echo "<script language=\"javascript\">alert('添加成功');history.go(-1)</script>"; }include("Head.php");?><SCRIPT language=javascript>function CheckPost(){ if (myform.user.value=="") { alert("用户名不能这空,请填写用户名"); myform.user.focus(); return false; } if (myform.title.value.length<5) { alert("标题不能少于5个字符"); myform.title.focus(); return false; } if (myform.content.value.length<8) { alert("内容不能少于8个字节"); myform.content.focus(); return false; } }</SCRIPT><form action="Textbook.php" method="post" name="myform" onsubmit="return CheckPost();"> <!--注：这里的action="Textbook.php"是定就是当前的文件名-->用户：<input type="text" size="10" name="user"><br>标题：<input type="text" name="title"><br>留言内容:<br><textarea name="content" rows=20 cols=30></textarea><br><input type="submit" name="submit" value="提交留言"></form> List.php文件代码：
<?include("Sql.php");include("Head.php"); $sql="SELECT * from message order by id desc"; //order by id desc 倒序排列 $query=mysql_query($sql); while($row=mysql_fetch_array($query)){?><table width=500 border="1" align="center" cellpadding="5" cellspacing="2" bgcolor="#add3ef"><tr bgcolor="#eff3ff"><td bgcolor ="eee3ff">ID序列:<?=$row[id]?></td> <td bgcolor ="eff3ff">标题:<?=$row[title]?></td> <td bgcolor ="#aeeffc">用户:<?=$row[user]?></td> <td bgcolor ="#ffffff">内容:<? echo htmtocode($row[content]);?></td></tr></table><? }?> 在纠正了“变量未定义”的警告后，未发现你说的现象
但不能排除是 sql.php 或 head.php 中的问题
变量未定义是不是指的这里
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