我有两个页面,一个将值发送到 php 页面的 html,然后,php 页面在 html 页面中以 div 形式返回结果。
在第一个测试中它只有一条错误消息,并且在 else 子句中它必须重定向到另一个 html 页面,因此 php 页面中的脚本将以 html 显示。我用过javascript,但它不适合我。
这是我使用的所有代码:
if (mysqli_num_rows($result)==0)
{
echo" Code erroné ";
}
else
{
header('location:infosInt.html');
//echo "<script> window.open('infosInt.html')</script>";
//echo "<script> window.location = 'infosInt.html'</script>";
}
编辑
我的html页面:
<!DOCTYPE html>
<head>
<script type="text/javascript" src="jquery.js"></script>
<script src="jquery-ui.js" charset="ISO-8859-1"></script>
<script src="jquery-1.9.1.js" charset="ISO-8859-1"></script>
<script type="text/javascript" src="jquery.crypt.js" ></script>
<LINK rel="stylesheet" type="text/css" href="android.css">
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<title>Login Form</title>
<script>
function connexion(evt)
{
var x=document.getElementById('code');
var cde = x.value;
var str = $().crypt({
method: "md5",
source: cde
});
if (cde=="")
{
// document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
var url ="http://localhost/filename/page.php";
var params="q="+ str;
xmlhttp.open("POST",url,true);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.setRequestHeader("Content-length", params.length);
xmlhttp.setRequestHeader("Connection", "close");
xmlhttp.send(params);
};
</script>
</head>
<body>
<div class="login">
<p></p>
<form method="" action="" id= "codeform" onsubmit="connexion();return false;">
<p>        </p>
<input type="password" id="code" ></p>
<p class="submit"><input type="submit" id="connect" value="Connexion" ></p>
</form>
<div id ="txtHint"></div>
</div>
</body>
</html>
这是page.php:
<?php
$host = "localhost";
$port = port;
$socket = "";
$user = "root";
$password = "";
$dbname = "base";
$con = new mysqli($host, $user, $password, $dbname, $port, $socket);
if (!$con)
{
die('Could not connect: ' . mysqli_error($con));
}
$q =$con->real_escape_string($_POST['q']);
mysqli_select_db($con,"ajax_demo");
$sql="query where column ='".$q."'"; // ex
$result = mysqli_query($con,$sql);
if (mysqli_num_rows($result)==0)
{
echo" Code erroné ";
}
else
{
echo"<head><meta http-equiv='refresh' content='5; URL=infosInt.html'></head>";}
mysqli_close($con);
?>