Question

我正在使用ajax将数据发送到PHP脚本并在数据库中设置查询。我的代码：

HTML脚本 ：

<span id="btn_span_<?php echo $user_id ?>">
      <?php if ($online == 1) { ?>
      <button onclick="update_online(1,<?php echo $user_id; ?>,'btn_span_<?php echo $user_id ?>')" class="btn-custom-delete btn btn-status">is active</button>
      <?php } elseif ($online == 0) { ?>
      <button onclick="update_online(0,<?php echo $user_id; ?>,'btn_span_<?php echo $user_id ?>')" class="btn-custom-services btn btn-status">active
      </button>
      <?php } ?>
 </span>

JAVASCRIPT脚本 ：

function update_online(status, id, span_id) {
var settings = {
     "async": true,
     "crossDomain": true,
     "url": "script_edit_status.php?status=" + status + '&id=' + id + '&span_id=' + span_id,
     "method": "GET"
};

$.ajax(settings).done(function (response) {
      var obj = JSON.parse(response);
      var btn = document.getElementById(span_id);
      if (obj.status == "1") {
            btn.innerHTML = "<button onclick='update_online(1,obj.id,obj.span_id)' class='btn-custom-services btn btn-status'>active</button>";
      } else if (obj.status == "0") {
                btn.innerHTML = "<button onclick='update_online(0,obj.id,obj.span_id)' class='btn-custom-delete btn btn-status'>is active</button>";
      }
});
}

PHP脚本 ：

<?php
include_once "../db/connection.php";
$id = $_GET['id'];
$status = $_GET['status'];
$span_id = $_GET['span_id'];
try {
    if ($status == 1) {
        $sql_edit_status = "update api_user set online=0 where id='$id';";
    } elseif ($status == 0) {
        $sql_edit_status = "update api_user set online=1 where id='$id';";
    }
    $conn->query($sql_edit_status);


    $status_arr = array(
        "id" => $id,
        "status" => $status,
        "span_id" => $span_id
    );

    echo json_encode($status_arr);

} catch (PDOException $e) {
    echo "Error: " . $e->getMessage();
}
$conn = null;

可以执行第一个操作，但是当我想单击second time的按钮时，consol显示以下错误：

未捕获的ReferenceError：未定义obj

该错误是由于innerHTML无法正确发送方法而引起的...
你能帮我吗？

Answer 1

btn.innerHTML = "<button onclick='update_online(0," + obj.id + "," + obj.span_id + ")' class='btn-custom-delete btn btn-status'>is active</button>";

innerHtml关于响应的ajax方法

1 个答案: