如何通过jquery ajax发送不同的行id来通过单击复选框来更新另一个php页面中的表记录？

Question

When I click on both green checkboxes, just the latest table id send, the blow code are both checkbox and jQuery,

Checkbox:

<div class='md-checkbox has-success'>
    <input type='checkbox' id='submit'  value='$id' class='md-check'>
    <label for='submit'>
        <span></span>
        <span class='check'></span>
        <span class='box'></span>
        Submit
    </label>
</div>

jQuery:

$("#submit").change(function() {
    var submit=$("#submit").val();
    var remark=$("#remark").val();
    var submit_date=$("#submit_date").val();
    var string_data = 'id=' + id + '&remark=' + remark + '&submit_date=' + submit_date + '&submit=' + submit;
    alert(string_data);
    $.ajax({
        type: "POST",
        url: "estimation_qc_update.php",
        data: string_data,
        cache: false,
        success: function(html) {
            $("#estimation").html('');
            $("#estimation").html(html);
        }
    });
}); 

Answer 1

You can cope with this problem by applying common css classes and custom data attributes

e.g.

<td calss="btn-submit" data-target-id="unique-record-id-goes-here">submit</td>

now you can do the miracles!

</div>