Php 单击mysqli中的另一个按钮后显示按钮

Php 单击mysqli中的另一个按钮后显示按钮,php,button,mysqli,Php,Button,Mysqli,我在index.php中有以下代码: <?php require 'class.php';require_once './session.php';$conn = new db_class();$read = $conn->read();$data = [];while($fetch = $read->fetch_array()){ $data[] = $fetch;}?> <table class = "table table-bordered table-

我在index.php中有以下代码:

<?php require 'class.php';require_once './session.php';$conn = new db_class();$read = $conn->read();$data = [];while($fetch = $read->fetch_array()){ 
$data[] = $fetch;}?>
<table   class = "table table-bordered table-responsive ">
        <thead>
            <tr>
                <th>Segment</th>
                <th>Action</th>
            </tr>
        </thead>
        <tbody>
            <?php foreach($data as $fetch): ?>
                <tr>
        <input type="hidden" name="user_id" value="<?php echo $_SESSION['user_id']?>">
                    <td class="segment" contenteditable="true"><?= $fetch['segment'] ?></td>
                    <td class="text-center">
                        <button class = "btn btn-default action-btn" data-id="<?= $fetch['id'] ?>" data-action="update"> 
                            <span class = "glyphicon glyphicon-edit"></span> Update
                        </button> 
                        | 

                        <button class = "btn btn-success action-btn" data-id="<?= $fetch['id'] ?>" type="submit" data-action="activate">
                            <span class = "glyphicon glyphicon-check"></span> Activate
                        </button> 
                        | 
                        <button class = "btn btn-danger action-btn" data-id="<?= $fetch['id'] ?>"  data-action="deactivate">
                            <span class = "glyphicon glyphicon-folder-close"></span> deactivate
                        </button>
                    </td>
                </tr>
            <?php endforeach; ?>

        </tbody>
    </table>
我只想显示是否禁用激活以仅显示激活按钮,如果启用激活以仅显示停用按钮,
我使用0表示停用,1表示启用

首先需要隐藏停用按钮使用
样式
显示
隐藏 按钮在activate按钮上写入click事件,然后使用jquery隐藏activate按钮并显示deactivate按钮

$(“.activate btn”)。单击(函数(){
$(this.hide();
$(“.deactivate btn”).show();
});

段
行动

您如何确定该按钮是否已激活。db有什么信息吗?是的,我有一个名为segment的表,其中defaut激活为1,因此我只想显示deactivate按钮,单击它后,我想显示activate按钮@Omi
  public function read()
{
    $stmt = $this->conn->prepare("SELECT id,segment FROM `segments`") or die($this->conn->error);
    if ($stmt->execute()) {
        $result = $stmt->get_result();
        return $result;
    }

}