Question

我正在cordova制作一个android应用，仅供我练习。该应用程序在图表上显示数据。 x axis dateTime和y axis中包含Energy个值。我知道这一定是一个重复的问题。但作为php和mysql中的新手，我想要创建一个jSON对象，并且在该对象中我想创建一个array。

现在我按照我的要求制作了一个hard-coded jSON对象并执行了我的任务。下面是我的json对象

var jsonData = {
"11111111": [
    { "x": "2016-01-25 12:58:52", "y": 25.22 },
    { "x": "2016-01-26 13:33:23", "y": 30.14 },
    { "x": "2016-01-27 13:49:18", "y": 35.58 },
    { "x": "2016-01-29 13:55:01", "y": 40.25 },
    { "x": "2016-01-30 14:00:15", "y": 47.25 },
    { "x": "2016-01-31 15:50:15", "y": 14.25 },
    { "x": "2016-02-01 16:16:15", "y": 20.25 },
    { "x": "2016-02-02 17:49:15", "y": 5.25 },
    { "x": "2016-02-03 18:20:15", "y": 77.25 },
    { "x": "2016-02-04 19:30:15", "y": 65.25 }
],
"22222222": [
     { "x": "2016-02-25 14:23:31", "y": 17.99 },
     { "x": "2016-02-26 14:30:36", "y": 27.78 },
     { "x": "2016-02-27 14:34:23", "y": 37.45 },
     { "x": "2016-02-28 14:42:47", "y": 47.73 },
     { "x": "2016-02-29 15:07:26", "y": 57.58 },
     { "x": "2016-03-01 16:17:26", "y": 32.58 },
     { "x": "2016-03-02 17:27:26", "y": 42.58 },
     { "x": "2016-03-03 18:40:26", "y": 62.58 },
     { "x": "2016-03-04 19:47:26", "y": 77.58 },
     { "x": "2016-03-05 20:37:26", "y": 67.58 }
],
"33333333": [
    { "x": "2016-03-25 15:14:49", "y": 15.66 },
    { "x": "2016-03-26 15:32:51", "y": 25.68 },
    { "x": "2016-03-27 15:40:32", "y": 35.73 },
    { "x": "2016-03-28 15:58:07", "y": 45.46 },
    { "x": "2016-03-29 16:21:25", "y": 46.79 },
    { "x": "2016-03-30 17:21:25", "y": 26.79 },
    { "x": "2016-04-01 18:21:25", "y": 56.79 },
    { "x": "2016-04-02 19:21:25", "y": 46.79 },
    { "x": "2016-04-03 20:22:25", "y": 60.79 },
    { "x": "2016-04-04 21:52:25", "y": 77.79 }
],
"44444444": [
    { "x": "2016-04-25 12:58:52", "y": 20.22 },
    { "x": "2016-04-26 13:33:23", "y": 30.14 },
    { "x": "2016-04-27 13:55:01", "y": 40.25 },
    { "x": "2016-04-28 14:23:31", "y": 50.99 },
    { "x": "2016-04-29 14:30:36", "y": 60.78 },
    { "x": "2016-04-30 15:35:36", "y": 45.78 },
    { "x": "2016-05-01 16:48:36", "y": 63.78 },
    { "x": "2016-05-02 17:58:36", "y": 77.78 },
    { "x": "2016-05-03 18:30:36", "y": 37.78 },
    { "x": "2016-05-04 19:30:36", "y": 49.78 }
],
"55555555": [
    { "x": "2016-05-25 14:42:47", "y": 55.73 },
    { "x": "2016-05-26 15:07:26", "y": 65.58 },
    { "x": "2016-05-27 15:14:49", "y": 75.66 },
    { "x": "2016-05-28 15:32:51", "y": 85.68 },
    { "x": "2016-05-29 16:10:07", "y": 95.46 },
    { "x": "2016-05-30 18:48:07", "y": 55.46 },
    { "x": "2016-06-01 19:18:07", "y": 48.46 },
    { "x": "2016-06-02 20:58:07", "y": 68.46 },
    { "x": "2016-06-03 21:28:07", "y": 105.46 },
    { "x": "2016-06-04 24:18:07", "y": 45.46 }
]
}

我想使用MySQL结果创建完全相同的副本。在上面的json中，11111111, 22222222, 33333333, 44444444, 55555555是meter的序列号，如上所述，x和y具有日期时间和能量值。序列号，日期时间和能量在同一个表中，序列号对于每个meter是唯一的。此外，这些序列号具有唯一ID但我不想使用它，因为我的搜索基于serial numbers而不是ID。

我的数据库也是实时的，即它在服务器上。下面是我的php代码，其中我创建了一个简单的表并显示了其中的所有数据。

<?php

  require_once('config.php');
  $sql = "SELECT Device_ID, Energy_kwh, Power_kW, Voltage_Phase_1,Voltage_Phase_2,Voltage_Phase_3, Current_Phase_1,Current_Phase_2,Current_Phase_3, Data_Datetime, Device_Serial_Number FROM ads_device_data where Device_ID= '2'";
  //$result = $con->query($sql);
  if($result = mysqli_query($con,$sql))
  {
    if(mysqli_num_rows($result)>0)
  {
    echo"<table id='myTable'>";
    echo"<tr>";
     echo"<th>ID</th>";
     echo"<th>Energy</th>";
     echo"<th>Power</th>";
     echo"<th>Voltage (all phases)</th>";
     echo"<th>Current (all phases)</th>";
     echo"<th>Date Time</th>";
     echo"<th>Serial #</th>";
    echo"</tr>";
    while($row = mysqli_fetch_array($result))
    {
        echo"<tr  style='text-align:center'>";
        echo"<td>" . $row['Device_ID'] . "</td>";
        echo"<td>" . $row['Energy_kwh'] . "</td>";
        echo"<td>" . $row['Power_kW'] . "</td>";
        echo"<td>" . $row['Voltage_Phase_1'] . " : " . $row['Voltage_Phase_2'] . " : " . $row['Voltage_Phase_3'] . "</td>";
        echo"<td>" . $row['Current_Phase_1'] . " : " . $row['Current_Phase_2'] . " : " . $row['Current_Phase_3'] . "</td>";
        echo"<td>" . $row['Data_Datetime'] . "</td>";
        echo"<td>" . $row['Device_Serial_Number'] . "</td>";
        echo"</tr>";
        //echo "id: ".$row["Device_ID"]." <br> "." -Energy ".$row["Energy_kwh"]." <br> "." -Power " .$row["Power_kW"]." <br> "." -Voltage(all phases) ".$row["Voltage_Phase_1"]. " " .$row["Voltage_Phase_2"]. " " .$row["Voltage_Phase_3"]." <br> ". "-Current(all phases)" .$row["Current_Phase_1"]. " " .$row["Current_Phase_2"]. " " .$row["Current_Phase_3"]." <br> ". " -Date Time " .$row["Data_Datetime"]." <br> "." Serial# ".$row["Device_Serial_Number"]. " <br> ";
    }
    echo"</table>";
     mysqli_free_result($result);
}
else
{
    echo "No records matching your query were found.";
}
}else
{
//$result->close();
//mysqli_free_result($result);
    echo "ERROR: Could not able to execute the query. " .   mysqli_error($con);
 }
  mysqli_close($con);
 //$con->close();
 ?>

任何帮助都将受到高度赞赏。

Answer 1

试试这个

$sql = "SELECT Device_ID, Energy_kwh, Power_kW, Voltage_Phase_1,Voltage_Phase_2,Voltage_Phase_3, Current_Phase_1,Current_Phase_2,Current_Phase_3, Data_Datetime, Device_Serial_Number FROM ads_device_data where Device_ID= '2'";
$result = mysqli_query($con, $sql);
$data = [];
if ($result) {
    while ($row = mysqli_fetch_array($result)) {
        $serial = $row['Device_Serial_Number'];
        $x = $row['Data_Datetime'];
        $y = $row['Energy_kwh'];

        if (!isset($data[$serial])) {
            $data[$serial] = [];
        }
        $data[$serial][] = ['x' => $x, 'y' => $y];
    }
    mysqli_free_result($result);
}
$json = json_encode($data, JSON_PRETTY_PRINT);
echo "var jsonData = " . $json;

来自PHP MySQL jSon对象

1 个答案: