來自PHP的jSon對象MySQL結果

Question

我在cordova上工作，製作android應用程序，僅供我練習。該應用在圖表上顯示數據。 x axis有dateTime和y axis其中有Energy值。 我知道這一定是一個重複的問題。但作爲新手在php和mysql我想創建一個jSON對象，並在該對象中我想創建一個array。

現在我已經根據我的要求製作了hard-coded jSON對象並執行了我的任務。下面是我的JSON對象

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 } 
] 
} 

我想用MySQL結果創造了完全相同的副本。 在上面的json中，11111111, 22222222, 33333333, 44444444, 55555555是我的序列號meter，如上所述x和y具有日期時間和能量值。序列號，日期時間和能量在同一個表中，序列號對於每個meter都是唯一的。此外，這些序列號具有唯一的ID，但我不想使用它，因爲我的搜索是基於serial numbers而不是ID。

另外我的數據庫是活的，即它在服務器上。以下是我的php代碼，其中我創建了一個簡單的表格並顯示其中的所有數據。

<?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(); 
?> 

任何幫助將不勝感激。

Answer 1

試試這個

$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;