我有以下内容用于更新DataTable。它更新我的数据库就好了但是在AJAX中的onSuccess它没有刷新表。我成功使用:$('#data-table').html(data);
但它无法正常工作。欢迎任何建议,提前谢谢!
我使用网站上的DataTables:https://datatables.net/。
我的PHP:
<?php
$connect = mysqli_connect("localhost", "root", "pass", "table");
if (!empty($_POST)) {
$output = '';
$message = '';
$numeclub = mysqli_real_escape_string($connect, $_POST["numeclub"]);
$coregraf = mysqli_real_escape_string($connect, $_POST["coregraf"]);
$numeformatie = mysqli_real_escape_string($connect, $_POST["numeformatie"]);
$disciplina = mysqli_real_escape_string($connect, $_POST["disciplina"]);
$categorievarsta = mysqli_real_escape_string($connect, $_POST["categorievarsta"]);
$sectiuni = mysqli_real_escape_string($connect, $_POST["sectiuni"]);
$numecoregrafie = mysqli_real_escape_string($connect, $_POST["numecoregrafie"]);
if ($_POST["id"] != '') {
$query = "UPDATE momente SET numeclub='$numeclub', coregraf='$coregraf',
numeformatie='$numeformatie', disciplina = '$disciplina',
categorievarsta = '$categorievarsta', sectiuni = '$sectiuni',
numecoregrafie = '$numecoregrafie' WHERE id='" . $_POST["id"] . "'";
$message = 'Data Updated';
}
if (mysqli_query($connect, $query)) {
$output .= '<label class="alert alert-success">' . $message . '</label>';
$select_query = "SELECT * FROM momente ORDER BY id DESC";
$result = mysqli_query($connect, $select_query);
}
echo $output;
}
这里的AJAX代码工作正常,但成功后不会刷新我的DataTable ...
$('#insert_form').on("submit", function (event) {
event.preventDefault();
$.ajax({
url: "update.php",
method: "POST",
data: $('#insert_form').serialize(),
beforeSend: function () {
$('#insert').val("Introduc datele");
},
success: function (data) {
$('#insert_form')[0].reset();
$('#add_data_Modal').modal('hide');
$('#data-table').html(data);
}
});
});