一個div我有有PHP代碼一個div(它從一個數據庫打印數據)。我做了一個AJAX功能,將數據發送到服務器而無需刷新頁面,下面的代碼:刷新AJAX的成功
var url = "loadsave.php";
$('.create_save').submit(function(event) {
if ($('.gss_new_save').val()==""){
event.preventDefault();
} else {
$.ajax({
type: "POST",
url: url,
data: $('form').serialize(),
success: function()
{
}
});
event.preventDefault();
}
});
這是打印的代碼的另一部分。問題是,它打印的代碼只是當你刷新頁面:
<?php
$connect = new PDO('mysql:host=localhost;dbname=game','root','');
$cmd = $connect->query('SELECT * FROM saves');
$result = $cmd -> fetchAll(PDO::FETCH_ASSOC);
foreach ($result as $res) {
?>
<div class="gss_save cls">
<span class="gss_save_text cls"><?php echo $res['name'];?></span>
<span class="gss_level">lvl: <?php echo $res['level'];?></span>
</div>
<?php
}
?>
編輯:我補充一下:形式:
<form method="post" action="loadsave.php" class="create_save">
<input type="text" class="gss_new_save" name="name">
<input type="submit" style="display:none">
</form>
和loadsave.php的內容:
<?php
if (isset($_POST['name'])) {
$nombre = $_POST['name'];
}
$connect = new PDO('mysql:host=localhost;dbname=game','root','');
$stm = $connect->query("SELECT * FROM saves WHERE `name` ='$nombre'");
$result = $stm->fetchAll(PDO::FETCH_ASSOC);
if(count($result)==0){
$cmd = $connect->prepare('INSERT INTO saves (name, level) VALUES (:nombre,"1")');
$cmd -> bindParam(":nombre",$nombre);
$cmd -> execute();
}
?>
它的功能就像一個魅力,但我想刷新div打印我剛剛發送的內容,我該如何實現?
'...我只是sent',從發送的內容,其中,服務器或形式你發送*到*服務器? – adeneo
表單就是它.. – pedritoalfonso
你想把整個表單放在DIV中嗎? – adeneo