Question

我有多个选择选项。但是，我想要当我选择1个选项或两个或其他选项以动态更改html的列<th></th>并从数据库获取数据

当前，我已经创建了多选选项和html表

<select class="selectpicker" id="sensor" name="sensor[]" multiple>
                                <option 
     value="temperature">Temperature</option>
                                <option value="humidity">Humidity</option>
                                <option value="presure">Presure</option>
                                <option 
     value="level">Level</option>
                                </select>

     $query = "SELECT ".$selectedSensorOption." from p1";       
                    $result = $db_handle->runQuery($query);
    foreach ($result as $key => $value) {
    <table>
    <tr>
    <th>$result</th>
    </tr> 
    <tr>
    <td>$result</td>

我也希望基于多个选择选项从MySQL数据库中获取动态列和字段

Answer 1

一种简单的方法是使用结果的键值作为标题，然后将每一行数据作为值输出（代码中的注释）...

// Output table start and header row
echo "<table><tr>";
// Use the first rows results as the header values
foreach ( $result[0] as $header => $value ) {
    echo "<th>{$header}</th>";
}
echo "</tr>";
// Output each data row
foreach ($result as $values) {
    echo "<tr>";
    // Loop over each item in the row and output the value
    foreach ( $values as $value ) {
        echo "<td>{$value}</td>";
    }
    echo "</tr>";
}
echo "</table>";

您可以创建较短的代码（使用implode()等），但是有时更简单的代码更容易开始和维护。

使用返回的列名意味着您可以使用带有列别名的特定标题-类似于

id as `User ID`, fullname as `Name`

Answer 2

我用过AJAX。希望对您有所帮助； D

read.php
<?php
    //include header
    header('Content-Type: application/json');
$conn= mysqli_connect("localhost","my_user","my_password","my_db");
if (mysqli_connect_errno()){
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$type = $_GET["type"];
if($type == "GetCategory"){
    //SAMPLE QUERY
    $sql = "SELECT CategoryID,CategoryName from Category";
    $data = array();
    $results = $db -> query($sql);
    while($row = mysqli_fetch_assoc($results)){
        $data[] = $row;
    }
    echo json_encode($data);
}
if($type == "GetSubCategory"){
    $CategoryID = $_POST["CategoryID"];
    //SAMPLE QUERY
    //LET'S ASSUME THAT YOU HAVE FOREIGN KEY
    $sql = "SELECT SubCategoryID,SubCategoryName from SubCategory  where CategoryID = '".$CategoryID."'";
    $data = array();
    $results = $db -> query($sql);
    while($row = mysqli_fetch_assoc($results)){
        $data[] = $row;
    }
    echo json_encode($data);
}
?>


index.html
<select id="category"></select>
<select id="subcategory"></select>

<!--PLEASE INCLUDE JQUERY RIGHT HERE E.G. <script src='jquery.min.js'></script>--> 
<!--DOWNLOAD JQUERY HERE https://jquery.com/-->
<script>
    LoadCategory();
    function LoadCategory(){
        $.ajax({
            url:"read.php?type=GetCategory",
            type:"GET",
            success:function(data){
                var options = "<option selected disabled value="">Select 
Category</option>";
            for(var i in data){
                options += "<option value='"+data[i].CategoryID+"'>" + data[i].CategoryName + "</option>";
            }
            $("#category").html(options);
        }
    });
}
function LoadSubCategory(CategoryID){
    $.ajax({
        url:"read.php?type=GetSubCategory",
        type:"POST",
        data:{
            CategoryID : CategoryID
        },
        success:function(data){
            var options = "<option selected disabled value="">Select Sub-Category</option>";
            for(var i in data){
                options += "<option value='"+data[i].SubCategoryID+"'>" + data[i].SubCategoryName + "</option>";
            }
            $("#subcategory").html(options);
        }
    });
}
$("#category").change(function(){
    LoadSubCategory(this.value);
});

Answer 3

首先，您必须获取标题，可以使用explode获取标题键。您将使用两个循环，第一个循环用于构建标头，第二个循环用于构建主体。在主体循环中，您必须使用另一个循环才能为每一列获取正确的值


<select class="selectpicker" id="sensor" name="sensor[]" multiple>
    <option value="temperature">Temperature</option>
    <option value="humidity">Humidity</option>
    <option value="presure">Presure</option>
    <option value="level">Level</option>
</select>
<?php
$query = "SELECT ".$selectedSensorOption." from p1";
$results = $db_handle->runQuery($query);
//get the headers if $selectedSensorOption is not an array
$headers = explode(',', $selectedSensorOption)
?>
<table>
    <thead>
        <tr>
            <!-- loop and fill the header -->
            <?php foreach ($headers as $header) : ?>
                <th><?= ucfirst($header) ?></th>
            <?php endforeach; ?>
        </tr>
    </thead>
    <tbody>
        <!-- loop and fill the body-->
        <?php foreach ($results as $result) : ?>
               <tr>
                  <!-- loop and fill each column -->
                   <?php foreach ($headers as $header) : ?>
                     <th><?= $result[$header] ?></th>
                   <?php endforeach; ?>
               </tr>
        <?php endforeach; ?>
    </tbody>
</table>

如何使html表列<th> </th>在mysql中基于选定的列动态更改

3 个答案: