我必须传递将信息插入数据库表后生成的Json数据，现在我必须将此信息传递给其他页面 BookingConformation.html 。
例如我们将获得表单服务器欢迎Mr / Mrs Mittel感谢预订家庭服务您的预订ID为1。所以请告诉我如何在javascript中传递获取表单服务器的信息我必须将此信息传递给其他页面，请帮助我。
脚本
<script>
$(document).ready(function(){
$("#register-form").validate({
rules: {
userName: "required",
email: {
required: true,
email: true
},
userContactNumber: "required"
},
messages: {
userName: "Please enter your Name",
userContactNumber: "Please enter your Mobile number",
email: "Please enter a valid email address",
},
submitHandler: function(form) {
var uName = $('#userName').val();
var mailId = $('#email').val();
var mobNum = $('#userContactNumber').val();
$.ajax({
url:"http://localhost/bookRoom/book.php",
type:"POST",
dataType:"json",
data:{type:"booking", Name:uName, Email:mailId, Mob_Num:mobNum},
ContentType:"application/json",
success: function(response){
//alert(JSON.stringify(response));
//alert("success");
alert(response);
var globalarray = [];
globalarray.push(response);
window.localStorage.setItem("globalarray", JSON.stringify(globalarray));
window.location.href = 'BookingConformation.html';
},
error: function(err){
window.location.href = 'error.html';
}
});
return false; // block regular submit
}
});
});
</script>
服务器代码
<?php
header('Access-Control-Allow-Origin: *');//Should work in Cross Domaim ajax Calling request
mysql_connect("localhost","root","7128");
mysql_select_db("service");
if(isset($_POST['type']))
{
if($_POST['type']=="booking"){
$name = $_POST ['Name'];
$mobile = $_POST ['Mob_Num'];
$mail = $_POST ['Email'];
$query1 = "insert into customer(userName, userContactNumber, email) values('$name','$mobile','$mail')";
$query2 = "insert into booking(cust_email, cust_mobile, cust_name) values('$mail','$mobile','$name')";
$result1=mysql_query($query1);
$result2=mysql_query($query2);
$id=mysql_insert_id();
$value = "Welcome Mr/Mrs ".$name." Thanks for booking home services your booking id is = ".$id;
echo json_encode($value);
}
}
else{
echo "Invalid format";
}
?>
BookingConformation.html
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript">
function myFunction() {
var globalarray = [];
var arrLinks =[];
arrLinks = JSON.parse(window.localStorage.getItem("globalarray"));
document.getElementById("booking").innerHTML = arrLinks;
}
</script>
</head>
<body>
<p id="booking" onclick="myFunction()">Click me to change my HTML content (innerHTML).</p>
</bod
Y'GT;
答案 0 :(得分：1)
如果这是你真正的服务器端代码那么......它完全不安全。您永远不应将用户发布的变量直接传递给您的查询。
$query2 = "insert into booking(cust_email, cust_mobile, cust_name) values('$mail','$mobile','$name')";
至少使用“mysql_real_escape_string”转义值，或使用预准备语句。并且......不再使用mysql，使用mysqli，它几乎与您使用的相同，但不会很快弃用。
另外，你是json编码一个不需要json编码的字符串，它只是一段文本而不是有效的json代码。这可能就是为什么@SimarjeetSingh Panghlia的答案对你不起作用的原因。
而不是json_encoding该值，编码结构化数组。
$response = array( "status" => true );
if(isset($_POST['type']))
{
if($_POST['type']=="booking"){
$name = mysql_real_escape_string( $_POST ['Name'] ));
$mobile = mysql_real_escape_string($_POST ['Mob_Num']);
$mail = mysql_real_escape_string($_POST ['Email']);
$query1 = "insert into customer(userName, userContactNumber, email) values('$name','$mobile','$mail')";
$query2 = "insert into booking(cust_email, cust_mobile, cust_name) values('$mail','$mobile','$name')";
$result1 = mysql_query($query1);
$result2 = mysql_query($query2);
$id = mysql_insert_id();
$response["message"] = "Welcome Mr/Mrs ".$name." Thanks for booking home services your booking id is = ".$id;/* make sure you strip tags etc to prevent xss attack */
}
}
else{
$response["status"] = false;
$response["message"] = "Invalid format";
}
echo json_encode($response);
/* Note that you are making the query using ContentType:"application/json", */
这意味着无论查询是否成功，您都应该使用json进行响应。我还建议使用一个名为jStorage的简单jQuery插件，它允许轻松获取/设置对象而无需序列化它们。
答案 1 :(得分：0)
你可以试试这个
在第一页声明变量
<script type="text/javascript">
var globalarray = [];
globalarray.push(response.d);
window.localStorage.setItem("globalarray", JSON.stringify(globalarray));
</script>
在第二页上调用该变量
<script type="text/javascript">
var globalarray = [];
var arrLinks =[];
arrLinks = JSON.parse(window.localStorage.getItem("globalarray"));
console.log(arrLinks);
</script>
答案 2 :(得分：0)
您可以使用sessionStorage来存储和检索JSON数据。
var complexdata = [1, 2, 3, 4, 5, 6];
// store array data to the session storage
sessionStorage.setItem("list_data_key", JSON.stringify(complexdata));
//Use JSON to retrieve the stored data and convert it
var storedData = sessionStorage.getItem("complexdata");
if (storedData) {
complexdata = JSON.parse(storedData);
}
使用sessionStorage.clear();