Question

我正在制作记录温度和湿度的程序。我想在google仪表上实时输出数据。我正在使用php访问MySQL。我遇到的问题是我希望每5秒调用一次数据并更新仪表。测量仪只会更新我做错了什么？

<html>
  <head>
   <script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
   <script type="text/javascript">
      google.charts.load('current', {'packages':['gauge']});
      google.charts.setOnLoadCallback(drawChart);

      function drawChart() {

         var data = google.visualization.arrayToDataTable([
          ['Label', 'Value'],
          ['TEMP', <?php
$servername = "localhost";
$username = "datalogger";
$password = "datalogger";
$dbname = "datalogger";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "SELECT temperature FROM logger ORDER BY datetime DESC LIMIT 1";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        echo $row["temperature"];
    }
} else {
    echo "0 results";
}

mysqli_close($conn);
?> ],
          ['Memory', 7],
          ['CPU', 55],
        ]);

        var options = {
          width: 400, height: 120,
          redFrom: 90, redTo: 100,
          yellowFrom:75, yellowTo: 90,
          minorTicks: 5
        };

        var chart = new google.visualization.Gauge(document.getElementById('chart_div'));

        chart.draw(data, options);

        setInterval(function() {
          data.setValue(0 ,1 , <?php
$servername = "localhost";
$username = "datalogger";
$password = "datalogger";
$dbname = "datalogger";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "SELECT temperature FROM logger ORDER BY datetime DESC LIMIT 1";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        echo $row["temperature"];
    }
} else {
    echo "0 results";
}

mysqli_close($conn);
?>
);
          chart.draw(data, options);
        }, 5000);
        setInterval(function() {
          data.setValue(1, 1, 40 + Math.round(60 * Math.random()));
          chart.draw(data, options);
        }, 5000);
        setInterval(function() {
          data.setValue(2, 1, 60 + Math.round(20 * Math.random()));
          chart.draw(data, options);
        }, 26000);
      }
    </script>
  </head>
  <body>
    <div id="chart_div" style="width: 400px; height: 120px;"></div>
  </body>
</html>

php google gauge数据可视化MySQL刷新问题。

0 个答案: