[英]how to display the result in same page?
這是我的代碼。 HTML
<html>
<head></head>
</body>
<form id="myform" action="formdata.php" method="post">
username:<input type="text" name="username" id="name"><br>
password:<input type="password" name="password" id="pass"><br>
firstname:<input type="text" name="firstname" id="fname"><br>
lastname:<input type="text" name="lastname" id="lname"><br>
<input type="submit" id="submit" value="register">
</form>
<div id="status_text"></div>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<script>
$("#submit").click(function(){
var username = $('#name').val();
var password = $('#pass').val();
var firstname = $('#fname').val();
var lastnamee = $('#lname').val();
var postData = '&username='+username+'&password='+password+'&firstname='+firstname+'&lastname='+lastname;
var status_text = $('#status_text');
//alert(postData);
//var mydata = {'username':name,'password':pass,'firstname':fname,'lastname':lname};
/*$.post($('#myform').attr('action'),
$('#myform:input').serializeArray(),
function(info){
$('status_text').html(info)
});*/
$.ajax({
url:"action",
type:"post",
success:function(info)
{
status_text.html(info);
}
});
});
</script>
</body>
</html>
這是我的數據庫PHP代碼
<?php
$servername = 'localhost';
$username = 'root';
$password = '';
$conn = new mysqli($servername, $username, $password);
$name = $_POST['username'];
$password = $_POST['password'];
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$sql = "INSERT INTO karthik.person(name,password,firstname,lastname)VALUES('$name','$password','$firstname','$lastname')";
if($conn->query($sql) === TRUE){
echo("record added success");
}else{
echo("failed".$conn->error);
}
$conn->close();
?>
每當我運行此代碼時,它會轉到下一頁並顯示“成功添加記錄”之類的結果,而不是我希望代碼在同一頁中顯示結果。 我幾乎嘗試了所有方法,但無法獲得預期的結果。
您需要告訴瀏覽器不要提交表單:
$("#submit").click(function(e) {
e.preventDefault(); // Prevents the default behaviour for the submit-action
//... the rest of your code
完成這項工作后,您應該開始修復@jeroen列表中的內容 。
您在這里遇到一些問題:
$_POST
變量。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.