Question

我有一个带有信息模式/按钮的选项，我想在选择更改时进行动态连接

我的代码就像这样

<select name="name_select" id="my_select">
  <option value="1">Rossi</option>
  <option value="2">Skunk</option>
  <option value="3">Ceres</option>
</select>

<button type="button" class="btn btn-primary btn-lg" data-toggle="modal" data-target="#myModal"> Information </button>

<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
  <div class="modal-dialog" role="document">
    <div class="modal-content">
      <div class="modal-header">
        <button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span></button>
        <h4 class="modal-title" id="myModalLabel">Personal information</h4>
      </div>
      <div class="modal-body">

// query mysql
<?php (...) SELECT name, address, phone FROM table WHERE id = #my_select ?>

<input type="text" name="name" value="<?php echo $name ?>">
<input type="text" name="address" value="<?php echo $adress ?>">
<input type="text" name="phone" value="<?php echo $phone ?>">

      </div>
      <div class="modal-footer">
        <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
        <button type="button" class="btn btn-primary">Save changes</button>
      </div>
    </div>
  </div>
</div>

如何实现jquery post / get来解决这个问题？

Answer 1

尝试使用ajax

此页面的脚本将是：

$(function () {
    $(document).on("change", "#my_select", function () {
        $.ajax({
            url: 'path/to/ajaxfile.php',
            type: 'GET',
            dataType: 'json',
            data: {id: $(this).val()},
        })
        .done(function(response) {
            if (response.status) {
                $("#myModal").find(".modal-body").html(response.html);
            } else {
                alert(status.message);
            }
        })
        .fail(function(data) {
            alert("Something went wrong please try again later.");
            console.log(data.responseText);
        });

    });
});

和/ ajaxfile.php的路径代码：

/**
 * Connect to the database and do other initialization if required.
 *
 * Assuming Procedual MySQLi.
 */

if (empty($_GET['id'])) {
    echo json_encode(array("status" => false, "message" => "There is no User Selected."));
    exit;
}

$id = mysqli_real_escape_string($connection, $_GET['id']);

$query = "SELECT name, address, phone FROM table WHERE id = '{$id}'";
$result = mysqli_query($connection, $query);

if (mysqli_num_rows($result) > 0) {
    $row = mysqli_fetch_assoc($result);
    $inputs = '<input type="text" name="name" value="'. $row['name'] .'">
        <input type="text" name="address" value="'. $row['address'] .'">
        <input type="text" name="phone" value="'. $row['phone'] .'">';

    echo json_encode(array("status" => true, "html" => $inputs));
    exit;
} else {
    echo json_encode(array("status" => false, "message" => "There is no user with that id."));
    exit;
}

我希望这会有所帮助。

选择onchange update php id然后查询mysql

1 个答案: