如何将 PHP 数据从数据库发送回 JQuery 并插入到表中

Posted 2023-05-09

【中文标题】如何将 PHP 数据从数据库发送回 JQuery 并插入到表中

【英文标题】：How to send PHP data from database back to JQuery and insert to a Table

【发布时间】：2020-09-03 01:25:48

【问题描述】：

正如标题所说，我正在尝试根据用户输入显示来自数据库的数据并将其显示在 html 表中。简而言之，我希望用户输入过滤到数据库的查询请求，然后以表格格式显示结果。

我的问题是我不确定如何将查询结果返回给 JQuery，然后将该数据转换为表格以便在 HTML 中显示。

到目前为止，这一直是我的方法，但它不起作用。我已经使用常规 javascript 成功地做到了这一点。但是我想尝试使用 JQuery 来解决这个特殊问题，因为它可以进一步简化代码。

jQuery:

 var ourObj = ;

  ourObj.data = "Ranges";
  ourObj.arPoints = ["SID":TABLID,"T1": T1,"T2": T2,"H1": H1,"H2": H2,"P1": P1,"P2": P2,"C1": C1,"C2": C2,
                     "Y1": Y1,"Y2": Y2,"M1": M1,"M2": M2,"D1": D1,"D2": D2,"m1": m1,"m2": m2];

  $.ajax(
    type:"post",
    url:"GetQuery.php",
    data:"data" : JSON.stringify(ourObj),
    success: function(response)  

      $('SensorData').html(response);

    
  );

PHP

<?php
if (isset($_POST["data"])) 

    // Decode our JSON into PHP objects we can use
    $data = json_decode($_POST["data"]);

    $id = $data->arPoints[0]->SID;

    $T1 = $data->arPoints[1]->T1;
    $T2 = $data->arPoints[1]->T2;
    $H1 = $data->arPoints[2]->H1;
    $H2 = $data->arPoints[2]->H2;
    $P1 = $data->arPoints[3]->P1;
    $P2 = $data->arPoints[3]->P2;
    $C1 = $data->arPoints[4]->C1;
    $C2 = $data->arPoints[4]->C2;

    $Y1 = $data->arPoints[5]->Y1;
    $Y2 = $data->arPoints[5]->Y2;
    $M1 = $data->arPoints[6]->M1;
    $M2 = $data->arPoints[6]->M2;
    $D1 = $data->arPoints[7]->D1;
    $D2 = $data->arPoints[7]->D2;
    $m1 = $data->arPoints[8]->m1;
    $m2 = $data->arPoints[8]->m2;

    // echo "Sensore ID: " . $id;
    // echo "Temperature: " . $T1 . " : " . $T2;
    // echo "Humidity: " . $H1 . " : " . $H2;
    // echo "Pressure: " . $P1 . " : " . $P2;
    // echo "CO: " . $C1 . " : " . $C2;

    $conn = mysqli_connect('localhost', 'root', '', 'sensors');

    if(empty($Y1) || empty($Y2) || empty($M1) || empty($M2) || empty($D1) || empty($D2) || empty($m1) || empty($m2))

        $query = "SELECT * FROM sensors, sensorsdata WHERE sensors.SensorID = sensorsdata.SensorID 
        AND sensorsdata.SensorID = '$id' 
        AND sensorsdata.Temperature BETWEEN $T1 AND $T2
        AND sensorsdata.Humidity BETWEEN $H1 AND $H2
        AND sensorsdata.Air_Pressure BETWEEN $P1 AND $P2
        AND sensorsdata.Carbon_Monoxide BETWEEN $C1 AND $C2
        order by sensorsdata.Date Desc"; 
    
    else
        $time1= mktime(12,$m1,0,$M1,$D1,$Y1);
        $time1= date("Y-m-d h:i:s", $time1);

        $time2= mktime(12,$m2,0,$M2,$D2,$Y2);
        $time2= date("Y-m-d h:i:s", $time2);

        $query = "SELECT * FROM sensors, sensorsdata WHERE sensors.SensorID = sensorsdata.SensorID 
        AND sensorsdata.SensorID = '$id' 
        AND sensorsdata.Temperature BETWEEN $T1 AND $T2
        AND sensorsdata.Humidity BETWEEN $H1 AND $H2
        AND sensorsdata.Air_Pressure BETWEEN $P1 AND $P2
        AND sensorsdata.Carbon_Monoxide BETWEEN $C1 AND $C2
        AND sensorsdata.Date BETWEEN '$time1' AND '$time2'
        order by sensorsdata.Date Desc"; 
    

    // Get Result
    $result = mysqli_query($conn, $query);

    // Fetch Data
    while ($row = mysql_fetch_array($result))
    
       ?>
          <tr><td><?php echo $row['Temperature']?></td>
          <td><?php echo $row['Humidity']?></td>
          <td><?php echo $row['Air_Pressure']?></td>
          <td><?php echo $row['Carbon_Monoxide']?></td>
          <td><?php echo $row['Date']?></td></tr>;
        <?php   
    

    mysqli_close($conn);

?>

HTML 表格

  <table class="table">
                <thead>
                    <tr>
                        <th>Temperature</th>
                        <th>Humidity</th>
                        <th>Air Pressure</th>
                        <th>Carbon Monoxide</th>
                        <th>Date</th>
                    </tr>
                </thead>
                <tbody id="SensorData">

                </tbody>
            </table>

【问题讨论】：

我不完全确定你在问什么。您想要更好的方法还是上面的代码不能正常工作？

我已经编辑了问题描述以便更好地理解。但简而言之，我只想将我的 PHP 文件中的查询结果发送回 JQuery，然后将数据插入表中。

您可以尝试 console.logging 您的响应或查看来自您的 php 端点的数据是否显示在网络选项卡中？另外 $('SensorData') 应该是 $('#SensorData')

啊，总是小事，不过我还是设法解决了这个问题。再次感谢那个小错字，我总是想念那些。

【参考方案1】：

哇，好的，解决了这个问题，总是小事。

问题 #1：（感谢 Luke T.）将 $(‘SensorData’) 更改为 $(‘#SensorData’) 的简单问题

问题 #2：问题在于我如何浏览 PHP 文件中的每一行数据， 改成

while ($row = mysql_fetch_array($result))
    
        ?>
        <tr><td><?php echo $row['Temperature']?></td>
        <td><?php echo $row['Humidity']?></td>
        <td><?php echo $row['Air_Pressure']?></td>
        <td><?php echo $row['Carbon_Monoxide']?></td>
        <td><?php echo $row['Date']?></td></tr>;
      <?php   
     

到

foreach ($result as $row)
    
        ?>
        <tr><td><?php echo $row['Temperature']?></td>
        <td><?php echo $row['Humidity']?></td>
        <td><?php echo $row['Air_Pressure']?></td>
        <td><?php echo $row['Carbon_Monoxide']?></td>
        <td><?php echo $row['Date']?></td></tr>;
      <?php   
     

其余部分几乎保持不变，它现在在表格中显示请求的数据，没有问题。