对不起，如果这是重复问题，我确实尝试在互联网上找到答案并使用它.. 但我仍然可以完成它......这是我的问题...我想知道如何在使用无ID登录时显示名称？它就像用户使用他们的 no ID 登录后的欢迎屏幕。 这是我的代码，如果它搞砸了，很抱歉......我的登录名.php
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="1234"; // Mysql password
$db_name="vronline"; // Database name
$tbl_name="user_information"; // Table name
// Connect to server and select databse.
$con = mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$sql="SELECT * FROM $tbl_name WHERE user_id='$myusername' and user_password='$mypassword'" ;
$result=mysql_query($sql);
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){
// Register $myusername, $mypassword and redirect to file "login_success.php"
session_register("myusername");
session_register("mypassword");
$row=mysql_fetch_array($result);
if ($row['group_id']==0){
header("location:../user.php");
}
elseif ($row['group_id']==1) {
header("location:../admin.php");
}
//header("location:../menu.php");
}
else {
echo "Wrong Username or Password";
}
?>
这是我的用户.php
<?php
session_start();
if(!session_is_registered(myusername)){
header("location:../login_main.php");
}
?>
<?php
include ('php/dbconnect.php');
$matric = $_session['myusername'];
mysql_query("SELECT name FROM user_information WHERE user_id=".$_SESSION['myusername']);
?>
<html>
<head>
<title>user</title>
<link href="source/loginstyle.css" rel="stylesheet" type="text/css" media="all" />
<link href="source/kepala.css" rel="stylesheet" type="text/css" media="all" />
<link href="source/content.css" rel="stylesheet" type="text/css" media="all" />
<link href="source/menuBox.css" rel="stylesheet" type="text/css" media="all" />
<link href="source/bottomfoot.css" rel="stylesheet" type="text/css" media="all" />
<link href="source/accessibility_foundicons.css" rel="stylesheet" type="text/css" media="all"/>
<link href="source/general_foundicons.css" rel="stylesheet" type="text/css" media="all" />
<script src="source/Chart.js"></script>
</head>
<body>
<div>
<!--head-->
<div class= "kepala">
<h1>
<a href="#"><span class="foundicon-smiley"></span></a> <?php $row['name']; ?>
</h1>
</div>
<div class= "content">
<h4>content</h4>
<div class="menu">
<div class="menu_box_list">
<ul>
<li><a href="php/displayuseronly.php"><span class="foundicon1-person"></span> manage user<i class="manageuser"> </i><div class="clear"></div> </a></li>
<li><a href="userAddvehicle.php"><span> </span>vehicle manager <i class="vehicle manager"> </i><div class="clear"></div> </a></li>
<li><a href="#"><span class= "foundicon-page"></span> report <i class="Report"> </i><div class="clear"></div> </a></li>
<li><a href="#"><span class= "foundicon-settings"></span> log out<i class="settings"> </i><div class="clear"></div> </a></li>
</ul>
<div class="clear"></div>
</div>
</div>
<div class = "2nd_menu">
</div>
</div>
<div class= "Bfooter">
<h4></h4>
</div>
</div>
</body>
</head>
</html
我无法解决此代码
mysql_query（"从user_information中选择名称，其中 user_id=".$_SESSION['我的用户名']）;
如何使它显示为名称而不是user_id？对不起，我对此不太好对不起，我忘了添加这个我希望用户名显示在此处
<a href="#"<span></span></a> <?php $row['name']; ?>
每当我写的时候，我都无法在那里得到名字......它要么不显示任何内容，要么出错
你可能想试试这个：
$qry_sel = "SELECT name FROM user_information WHERE user_id=".$_SESSION['myusername'].";";
$qry_res = mysql_query($qry_sel);
$row = mysql_fetch_array($qry_res);
echo $row['name'];
在您的登录名.php页面中使用此代码
header("location:../user.php");
session_start();
$_SESSION['myusername']=$myusername;
并在用户中尝试此操作.php
ob_start("ob_gzhandler");
session_start();
$matric = $_session['myusername'];
$result=mysql_query("SELECT name FROM user_information WHERE user_id='$matric'");
if(mysql_num_rows($result)>0)
{
while($r=mysql_fetch_array($result))
{
$name=$r['name'];
echo $name;
}
}