Question

我有以下页面，它适用于MySQL，PHP和AJAX 如果我点击一个名称（id =＆＃34; orderN＆＃34;），它会返回咨询的结果，该结果命令名称降序或升序。有没有办法，如果你刷新（F5）页面，结果将保存为关闭前（ASC或DESC）？我听说过cookie和HTML5存储，这比cookie更好。如果你能用它们中的任何一个做，请告诉我

<html>
<head>
    <script type="text/javascript" src="jquery-1.8.2.min.js"></script>
</head>
<body>
    <table>
        <tr><th><a href="#" id="orderN">Name</a></th></tr> 
    </table>
    <?
    $Conn = mysql_pconnect('localhost', 'root', '1234') or die('Error"');
    mysql_select_db('DATA');
    $consult = "SELECT NAME
            FROM STUDENTS";
    $query = mysql_query($consult);
    echo "<div id='DivConsult'><table>";
    while ($table = mysql_fetch_assoc($query)) {
        echo "<tr>";
        echo "<td>" . $table['NAME'] . "</td>";
        echo "</tr>  ";}
    echo "</table>";
    ?>
    <script>
        $(document).ready(function() {
            var contName = 0;
            $('#orderN').click(function() {
                contName++;
                if (contName % 2 !== 0) {
                    $.ajax({
                        type: "POST",
                        url: "reOrder.php",
                        data: "tipOrder=ASC",
                        success: function(data) {
                            $('#DivConsult').html(data);
                        }});
                }
                if (contName % 2 == 0) {
                    $.ajax({
                        type: "POST",
                        url: "reOrder.php",
                        data: "tipOrder=DESC",
                        success: function(data) {
                            //alert(data);
                            $('#DivConsult').html(data);
                        }});
                }
            });
        });
    </script>
</body>

AJAX：

<?php
$Conn = mysql_pconnect('localhost', 'root', '1234') or die('Error"');
mysql_select_db('DATA');

$consult = "";

if (isset($_POST['tipOrder'])) {

 if ($_POST['tipOrder'] == 'ASC') {
    $consult = "SELECT NOMBRE
            FROM STUDENTS ORDER BY NAME ASC";
}
if ($_POST['tipOrder'] == 'DESC') {
    $consult = "SELECT NAME
            FROM STUDENTS ORDER BY NAME DESC";
}}`
  $query = mysql_query($consult);

echo "<table>";

while ($table = mysql_fetch_assoc($query)) {
    echo "<tr>";
    echo "<td>" . $table['Name'] . "</td>";
    echo "</tr>  ";}

echo "</table>";

?>

Answer 1

你可以做到但只是保存一个容器（任何div，span或甚至是body）

localStorage.variableName = document.getElementById("id");

然后您可以使用

进行访问

if(Storage!=="undefined" && localStorage.variableName!=null)

现在您可以将值设置为

container.val = localStorage.variableName

如何使用html5网络存储保存页面？

1 个答案: