////////////////// Jquery AJAX submit zonder te refreshen ///////////////////////////
function submitdata() {
var vakken = document.getElementById("vakken").value;
var dataString = 'vakken=' + vakken;
// AJAX
$.ajax({
type: "POST",
url: "prototype.php",
data: dataString,
cache: false,
success: function(html) {
alert(html);
}
});
return false;
}
這裏有什麼問題? ///////////////////////////////////////////////// /////////////獲取數據並無刷新頁面提交
<form id="selector" action="prototype.php" method="post" enctype="multipart/form-data">
<select name="vakken">
<option value="DED">VAK: DED </option>
<option value="UXU">VAK: UXU </option>
<option value="SCO">VAK: SCO </option>
<option value="PO">VAK: PO </option>
</select>
<button type="button" onClick="submitdata();">Submitii</button>
</form>
</div>
<div id="vlakkencontainer">
<?php
// recieve data
if(isset($_POST['vakken'])) {
$vak = $_POST['vakken'];
echo $vak;
$sqldedquery = mysqli_query($con, "SELECT * FROM `opdracht` WHERE `vak` = '" . $vak . "'");
$vlakid = 0;
while ($row = mysqli_fetch_array($sqldedquery)) {
$afbeelding = $row['afbeeldingnaam'];
$pdf = $row['pdfnaam'];
$naamopdracht = $row['opdrachtnaam'];
echo '<a id="vakhover" href="../../../db/' . "$pdf" . '"><div class="vlak" id=' . "$vlakid++" . '>
<img src="../../../db/' . $afbeelding . '"><div id="naamopdracht">' . $naamopdracht . '</div>
</div></a>';
}
} ?>
我怎樣才能從我的數據庫中獲取這些數據而不刷新頁面。我嘗試了Ajax,但我無法讓它工作!
什麼是你所得到的問題? –
當您使用AJAX從服務器獲取數據時,您應該在div或輸入中顯示該數據。所以你可以在服務器端構建HTML,而JS會在客戶端更新它而不刷新頁面。 – robot9706
它不起作用。當我按下按鈕時他會刷新,但它也不會顯示我的東西。 – rickkorsten