我正在設計酒店預訂系統,在插入客戶信息後,我必須發回帶有客戶信息插入時自動生成的booking_id的確認頁面。如何獲取插入表的行的自動生成的id?
我的問題是如何獲得在同一個PHP代碼預約ID數據的插入後
請注意還不止一個人插入數據,所以我認爲它無法實現通過剛剛返回插入的最後一行 請幫我解決這個問題。
我的JavaScript:
<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();
$.ajax({
url:"http://localhost/bookRoom/book.php",
type:"POST",
dataType:"json",
data:{type:"booking", Name:uName, Email:mailId, Mob_Num:mobNum},
ContentType:"application/json",
success: function(response){
window.location.href = 'BookingConformation.html';
},
error: function(err){
window.location.href = 'error.html';
}
});
return false; // block regular submit
}
});
});
</script>
我的服務代碼book.php中
<?php
header('Access-Control-Allow-Origin: *');//Should work in Cross Domaim ajax Calling request
mysql_connect("localhost","root","2190");
mysql_select_db("hotels");
if(isset($_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')";
$query2 = "insert into booking(cust_name, cust_email, cust_mobile) values('$name', '$mail','$mobile')";
$result1=mysql_query($query1);
$result2=mysql_query($query2);
echo json_encode($result1);
}
}
else{
echo "Invalid format";
}
?>
等什麼,現在是你的問題? – 2015-02-11 11:41:11
@CBroe,對不起,我更新請檢查它.. – geeks 2015-02-11 11:44:41
@noob,對不起,我更新請檢查它 – geeks 2015-02-11 11:45:06