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让我开始说我已经阅读了大约15个类似的线程在这里和其他网站试图解决我的问题发布之前。我试图用一些PHP和MySQL测试网站。一些信息：警告：mysqli_query（）期望参数1是mysqli，给出的字符串
我建立了本地托管的站点，然后导出到000 webhost。为了测试php和mysql的功能，这甚至是必要的吗？我本地有一个mysql数据库。这会造成问题吗？
我有两个问题。
- 在运行任何PHP脚本，我得到以下警告页面：
警告：虚拟（）已经被禁用在/home/a2575478/public_html/Login.php安全原因线2
警告：mysqli_query（）预计参数1是mysqli的，在/home/a2575478/public_html/Login.php空给出上线37
警告：mysqli_error（）预计参数1至被mysqli的，空在/home/a2575478/public_html/Login.php给出线37
我知道这些警告给mysql_ *函数关系被弃用，但我有麻烦固定弃用的功能
现在，当我尝试访问本地网页时，我得到的是“网页不可用”错误。是由于错误或只是试图在本地运行PHP？
该网站可达here
让我知道如果你需要任何更多的信息。
下面的代码。
的login.php
<?php @session_start(); ?>
<?php virtual('/Connections/logincon.php'); ?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "")
{
if (PHP_VERSION < 6) {
$theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
}
$theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);
switch ($theType) {
case "text":
$theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
break;
case "long":
case "int":
$theValue = ($theValue != "") ? intval($theValue) : "NULL";
break;
case "double":
$theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
break;
case "date":
$theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
break;
case "defined":
$theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
break;
}
return $theValue;
}
}
$query_Login = "SELECT * FROM `user`";
$Login = mysqli_query($logincon,$query_Login) or die(mysqli_error($logincon));
$row_Login = mysql_fetch_assoc($Login);
$totalRows_Login = mysql_num_rows($Login);
mysqli_select_db($database_logincon, $logincon);
?>
<?php
// *** Validate request to login to this site.
if (!isset($_SESSION)) {
session_start();
}
$loginFormAction = $_SERVER['PHP_SELF'];
if (isset($_GET['accesscheck'])) {
$_SESSION['PrevUrl'] = $_GET['accesscheck'];
}
if (isset($_POST['Username'])) {
$loginUsername=$_POST['Username'];
$password=$_POST['Password'];
$MM_fldUserAuthorization = "UserLevel";
$MM_redirectLoginSuccess = "/Account.php";
$MM_redirectLoginFailed = "/Login.php";
$MM_redirecttoReferrer = true;
mysql_select_db($database_logincon, $logincon);
$LoginRS__query=sprintf("SELECT Username, Password, UserLevel FROM `user` WHERE Username=%s AND Password=%s",
GetSQLValueString($loginUsername, "text"), GetSQLValueString($password, "text"));
$LoginRS = mysql_query($LoginRS__query, $logincon) or die(mysql_error());
$loginFoundUser = mysql_num_rows($LoginRS);
if ($loginFoundUser) {
$loginStrGroup = mysql_result($LoginRS,0,'UserLevel');
if (PHP_VERSION >= 5.1) {session_regenerate_id(true);} else {session_regenerate_id();}
//declare two session variables and assign them
$_SESSION['MM_Username'] = $loginUsername;
$_SESSION['MM_UserGroup'] = $loginStrGroup;
if (isset($_SESSION['PrevUrl']) && true) {
$MM_redirectLoginSuccess = $_SESSION['PrevUrl'];
}
header("Location: " . $MM_redirectLoginSuccess);
}
else {
header("Location: ". $MM_redirectLoginFailed);
}
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<link href="CSS/Layout.css" rel="stylesheet" type="text/css" />
<link href="CSS/Menu.css" rel="stylesheet" type="text/css" />
<link href="CSS/Menu.css" rel="stylesheet" type="text/css" />
<link href="/SpryAssets/SpryValidationTextField.css" rel="stylesheet" type="text/css" />
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<script src="/SpryAssets/SpryValidationTextField.js" type="text/javascript"></script>
</head>
<body>
<div id="Holder">
<div id="Header"></div>
<div id="NavBar">
<nav>
<ul>
<li><a href="Login.php">Login</a></li>
<li><a href="Register.php">Register</a></li>
<li><a href="ForgotPassword.php">Forgot Password</a></li>
</ul>
</nav>
</div>
<div id="Content">
<div id="PageHeading">
<h1>Home Page </h1>
</div>
<div id="ContentLeft">
<h2>This is a test page for a project in progress</h2>
<p> </p>
<h6>A work in progress. </h6>
</div>
<div id="ContentRight">
<form id="form1" name="form1" method="POST" action="<?php echo $loginFormAction; ?>">
<span id="sprytextfield1">
<label for="LoginForm"></label>
<span class="textfieldRequiredMsg">A value is required.</span></span>
<table width="400" border="0" align="center">
<tr>
<td><h6><span id="sprytextfield2">
<label for="Username"></label>
Username:<br />
<br />
<input name="Username" type="text" class="StyleTxtField" id="Username" />
</span></h6>
<span><span class="textfieldRequiredMsg">A value is required.</span></span></td>
</tr>
<tr>
<td> </td>
</tr>
<tr>
<td><h6><span id="sprytextfield3">
<label for="Password"></label>
Password:<br />
<br />
<input name="Password" type="text" class="StyleTxtField" id="Password" />
</span></h6>
<span><span class="textfieldRequiredMsg">A value is required.</span></span></td>
</tr>
<tr>
<td> </td>
</tr>
<tr>
<td><input type="submit" name="LoginButton" id="LoginButton" value="Login" /></td>
</tr>
<tr>
<td> </td>
</tr>
</table>
</form>
</div>
</div>
<div id="Footer"> </div>
</div>
<script type="text/javascript">
var sprytextfield1 = new Spry.Widget.ValidationTextField("sprytextfield1");
var sprytextfield2 = new Spry.Widget.ValidationTextField("sprytextfield2");
var sprytextfield3 = new Spry.Widget.ValidationTextField("sprytextfield3");
</script>
</body>
</html>
<?php
mysql_free_result($Login);
?>
Logincon.php，管理数据库连接
<?php
# FileName="Connection_php_mysql.htm"
# Type="MYSQL"
# HTTP="true"
$hostname_logincon = "172.16.1.72:3306";
$database_logincon = "logindb";
$username_logincon = "root";
$password_logincon = "boombox";
$logincon = mysql_pconnect($hostname_logincon, $username_logincon, $password_logincon) or trigger_error(mysql_error(),E_USER_ERROR);
?>
预先感谢任何答复
您正在混合驱动程序。 'Mysql'' ='mysqli'。例如'$ Login = mysqli_query（$ logincon，$ query_Login）'< - 'mysqli'然后...'$ row_Login = mysql_fetch_assoc（$ Login）;'你用'mysql_'然后同样的驱动' $ totalRows_Login = mysql_num_rows（$ Login）;'然后回到'mysqli'，'mysqli_select_db（$ database_logincon，$ logincon）;' – chris85
错误正好说明问题是什么 –
用'include（'不管怎么样，我都会使用'virtual（）'（禁用，因为错误信息中提到） –