Question

如代码实验室示例代码中所建议，我正在按以下方式实现报告状态：

<?php
// Get a connection for the database
require_once('mysqli_connect.php');

// Create a query for the database
$query = "SELECT username, message FROM messages";

// Get a response from the database by sending the connection
// and the query
$response = @mysqli_query($dbc, $query);

// If the query executed properly proceed
if($response){

echo '<table align="left"
cellspacing="5" cellpadding="8">

<tr><td align="left"><b>username</b></td>
<td align="left"><b>Message</b></td></tr>';

// mysqli_fetch_array will return a row of data from the query
// until no further data is available
while($row = mysqli_fetch_array($response)){

echo '<tr><td align="left">' . 
$row['username'] . '</td><td align="left">' . 
$row['message'] . '</td><td align="left">';

echo '</tr>';
}

echo '</table>';

} else {

echo "Couldn't issue database query<br />";

echo mysqli_error($dbc);

}

// Close connection to the database
mysqli_close($dbc);

?>

但这给了我以下答复：

<?php
DEFINE ('DB_USER', 'root');
DEFINE ('DB_PASSWORD', 'password');
DEFINE ('DB_HOST', 'localhost');
DEFINE ('DB_NAME', 'Cheese');

$dbc = @mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
OR die('Could not connect to MySQL: ' .
mysqli_connect_error());
?>

其中发布数据的值为：

function reportState(devid, actionVal) {
  console.log('INSIDE REPORT STATE');
  if (!app.jwt) {
    console.warn('Service account key is not configured');
    console.warn('Report state is unavailable');
    return;
  }




const postData = {
    requestId: 'hgrwbj', /* Any unique ID */
    agentUserId: '123', /* Hardcoded user ID */
    payload: {
      devices: {
        states: {
          [devid]: {
            on: actionVal,
          },
        },
      },
    },
  };



  console.log('POSTDATA %j', postData);

  return app.reportState(postData)
     .then((data) => {
       console.log('Report state came back');
       console.info(data);
     });
};

所以我尝试通过以下另一种方式实现它：

{
  "error": {
    "code": 400,
    "message": "Request contains an invalid argument.",
    "status": "INVALID_ARGUMENT"
  }
}

但是这次它只给出请求ID作为响应：

{"requestId":"hgrwbj","agentUserId":"123","payload":{"devices":{"states":{"0123456789:01":{"on":"true"}}}}}

这次Postdata是：

function reportState(devid, actionVal) {
  console.log('INSIDE REPORT STATE');
  if (!app.jwt) {
    console.warn('Service account key is not configured');
    console.warn('Report state is unavailable');
    return;
  }

  const jwtClient = new google.auth.JWT(
    jwt.client_email,
    null,
    jwt.private_key,
    ['https://www.googleapis.com/auth/homegraph'],
    null
  );
  console.log('JWT',jwt);
  console.log('JWTCLIENT',jwtClient);

  const postData = {
    requestId: 'hgrwbj', /* Any unique ID */
    agentUserId: '123', /* Hardcoded user ID */
    payload: {
      devices: {
        states: {
          [devid]: {
            on: actionVal,
          },
        },
      },
    },
  };

  jwtClient.authorize((err, tokens) => {
    if (err) {
      console.error(err);
      return;
    }
    console.log('ACCESS TOKEN',tokens.access_token);
    const options = {
      hostname: 'homegraph.googleapis.com',
      port: 443,
      path: '/v1/devices:reportStateAndNotification',
      method: 'POST',
      headers: {
        Authorization: ` Bearer ${tokens.access_token}`,
      },
    };
    return new Promise((resolve, reject) => {
      let responseData = '';
      const req = https.request(options, (res) => {
        res.on('data', (d) => {
          responseData += d.toString();
        });
        res.on('end', () => {
          resolve(responseData);
        });
      });
      req.on('error', (e) => {
        reject(e);
      });
      // Write data to request body
      req.write(JSON.stringify(postData));
      req.end();
    }).then((data) => {
      console.info('REPORT STATE RESPONsE', data);
    });
  });


  console.log('POSTDATA %j', postData);

};

关于我在哪里得到正确答案的任何建议？预先感谢。

Answer 1

在您的报告状态：

"on": "true"

您正在将属性“ true”作为字符串发送。服务器需要一个布尔类型，不带引号：

"on": true

应该可以正常工作。

实施Google Home报告状态时出错

1 个答案: