Php 使用jquery的AJAX POST请求

有人能帮我为这种情况创建一个ajax请求吗？我想如果我点击ok id，我只想更新它的内容并存储在数据库中。希望这里的任何人都能在这方面帮助我。多谢各位

查看文件：

<tbody>
    <?php foreach ($ds_name as $dsname): ?>
    <tr>
        <td><div class="text-infoo<?= $dsname->id ?>"><?= $dsname->name ?></div></td>
        <td>
            <a href="#" id="edit<?= $dsname->id ?>"><span class="glyphicon glyphicon-pencil"></span></a>
            <a href="#" id="ok<?= $dsname->id ?>"><span class="glyphicon glyphicon-ok"></span></a>
            <a href="#"><span class="glyphicon glyphicon-remove"></span></a>
        </td>
    </tr>
    <?php endforeach; ?>
</tbody>

<?php foreach ($ds_name as $dsname): ?>
    $('#edit<?= $dsname->id ?>').click(function () {
        var text = $('.text-infoo<?= $dsname->id ?>').text();
        var input = $('<input id="attribute<?= $dsname->id ?>" type="text" data-id="<?php echo $ds_content->id; ?>"value="' + text + '" />');
        $('.text-infoo<?= $dsname->id ?>').text('').append(input);
        input.select();

        input.blur(function () {
            var text = $('#attribute<?= $dsname->id ?>').val();
            $('#attribute<?= $dsname->id ?>').parent().text(text);
            $('#attribute<?= $dsname->id ?>').remove();
        });
        $('#edit<?= $dsname->id ?>').hide();
        $('#ok<?= $dsname->id ?>').show();
    });
<?php endforeach; ?>

你可以这样做

$.ajax({
            url: "data.php",//file wich has query select to db table
            data: {id:theid},//describe your data here
            dataType: 'json',// type of data that will you get (JSON/HTML).
            type: 'POST',//sending type (POST/GET)
            success: function(data) {
               //do change the select option
            }
        });

祝你好运