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讓我開始說我已經閱讀了大約15個類似的線程在這裏和其他網站試圖解決我的問題發佈之前。我試圖用一些PHP和MySQL測試網站。一些信息：警告：mysqli_query（）期望參數1是mysqli，給出的字符串
我建立了本地託管的站點，然後導出到000 webhost。爲了測試php和mysql的功能，這甚至是必要的嗎？我本地有一個mysql數據庫。這會造成問題嗎？
我有兩個問題。
- 在運行任何PHP腳本，我得到以下警告頁面：
警告：虛擬（）已經被禁用在/home/a2575478/public_html/Login.php安全原因線2
警告：mysqli_query（）預計參數1是mysqli的，在/home/a2575478/public_html/Login.php空給出上線37
警告：mysqli_error（）預計參數1至被mysqli的，空在/home/a2575478/public_html/Login.php給出線37
我知道這些警告給mysql_ *函數關係被棄用，但我有麻煩固定棄用的功能
現在，當我嘗試訪問本地網頁時，我得到的是「網頁不可用」錯誤。是由於錯誤或只是試圖在本地運行PHP？
該網站可達here
讓我知道如果你需要任何更多的信息。
下面的代碼。
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，管理數據庫連接
<?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, $password_logincon) or trigger_error(mysql_error(),E_USER_ERROR);
?>
預先感謝任何答覆
您正在混合驅動程序。 'Mysql'' ='mysqli'。例如'$ Login = mysqli_query（$ logincon，$ query_Login）'< - 'mysqli'然後...'$ row_Login = mysql_fetch_assoc（$ Login）;'你用'mysql_'然後同樣的驅動' $ totalRows_Login = mysql_num_rows（$ Login）;'然後回到'mysqli'，'mysqli_select_db（$ database_logincon，$ logincon）;' – chris85
錯誤正好說明問題是什麼 –
用'include（'不管怎麼樣，我都會使用'virtual（）'（禁用，因爲錯誤信息中提到） –