这两种登陆验证方法错哪了
这两种登陆验证方法哪里错了
第一种:
- PHP code
<!-- Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ --><?php $id=$_POST['id']; $password=$_POST['password']; $conn=mysql_connect("localhost","root","root"); if (!$conn){ die("连接失败:".mysql_errno()); } mysql_select_db("emp",$conn) or die(mysql_errno()); mysql_query("set names utf8")or die(mysql_errno()); $sql = "select count(id) from admin where id='".$id."' and password='".md5($password)."'"; $res = mysql_query($sql,$conn); $num = mysql_num_rows($res); if($num){ header("location:empManage.php"); exit(); }else{ header("location:login.php?errno=1"); exit(); } mysql_free_result($res); mysql_close($conn); ?>
此方法无论密码是否和数据库匹配都可以登陆
方法二:
- PHP code
<!-- Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ --><?php $id=$_POST['id']; $password=$_POST['password']; $conn=mysql_connect("localhost","root","root"); if (!$conn){ die("连接失败:".mysql_errno()); } mysql_select_db("emp",$conn) or die(mysql_errno()); mysql_query("set names utf8")or die(mysql_errno()); $sql="select password from admin where id=$id"; $res=mysql_query($sql,$conn); if ($row=mysql_fetch_assoc($res)){ if ($row['password']==md5($password)){ header("location:empManage.php"); } } header("location:login.php?errno=1"); mysql_free_result($res); mysql_close($conn); ?>
此方法即使id和密码都正确都无法登陆
求高人指点,看看哪里错了导致无法登陆
------解决方案--------------------
忘了还有第二个了。
逻辑问题我想
if ($row=mysql_fetch_assoc($res)){
if ($row['password']==md5($password)){
header("location:empManage.php");
}
}
header("location:login.php?errno=1");
假设成功,先header("location:empManage.php"),可到最后仍然会执行header("location:login.php?errno=1")。 header()函数并不会执行后立即发送消息报头给浏览器,于是后面的覆盖了前面的。不过没测试过所以不敢打保票。
把逻辑理顺了看看
if ($row=mysql_fetch_assoc($res)){
if ($row['password']==md5($password)){
header("location:empManage.php");
exit();
}
}
header("location:login.php?errno=1");
------解决方案--------------------
$num = mysql_num_rows($res);
if($num){
header("location:empManage.php");
这里错了,应该是 读取count值,然后判断count是否为1则可

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