将值传递到表格行单击的下一页？

Question

我有一个当前不可点击的表格。单击行时，我想从该表转到详细信息页面，并在过程中传递 id 值。

我知道我需要在表格行上创建一个 href 并以某种方式在点击时传递 id 值，但不清楚如何执行此操作。相关表和php：

  <div class="row">
                    <div class="col-12">
                        <div class="card">
                            <div class="card-body">
                                <div class="table-responsive m-t-40">
                                    <table id="example23" class="display nowrap table table-hover table-striped table-bordered" cellspacing="0" width="100%">
                                        <thead>
                                            <tr>
                                            <thead><tr><th title="Field #1">id</th>
                                            <th title="Field #2">Organization</th>
                                            <th title="Field #3">xxx</th>
                                            <th title="Field #4">xxx</th>
                                            <th title="Field #5">xxx</th>
                                            <th title="Field #6">xxx</th>
                                            </tr>
                                        </thead>
                                        <tbody>

                                            <?php

                                                $servername = "xxx";
                                                $username = "xxx";
                                                $password = "xxx";
                                                $dbname = "xxx";

                                                // Create connection
                                                $conn = new mysqli($servername, $username, $password, $dbname);

                                                // Check connection
                                                if ($conn->connect_error) {

                                                    die("Connection failed: " . $conn->connect_error);

                                                }else{

                                                    echo '<script>console.log("Connection successful!")</script>';

                                                }

                                                $SELECT = mysqli_query($conn,"SELECT * FROM `organization`");

                                                if($SELECT != false)
                                                {
                                                    while($rows = mysqli_fetch_array($SELECT))
                                                    {
                                                        echo "
                                                            <tr>
                                                                <td>".$rows["id"]."</td>
                                                                <td>".$rows["name"]."</td>
                                                                <td>".$rows["xxx"]."</td>
                                                                <td>".$rows["xxx"]."</td>
                                                                <td>".$rows["xxx"]."</td>
                                                                <td>".$rows["xxx"]."</td>
                                                            </tr>
                                                            ";
                                                    }
                                                }else{
                                                    echo "
                                                        <tr>
                                                        <td colspan='3'>Something went wrong with the query</td>
                                                        </tr>
                                                    ";
                                                }

                                            ?>

                                        </tbody>
                                    </table>
                                </div>
                            </div>
                        </div>
                    </div>
                </div>

如何使我的表格行可点击（以转到详细信息页面）以及如何将点击时表格中的 id 值传递到详细信息页面？

Answer 1

您需要制作href 标签并像这样传递id

<a href="paganame?id=<?php echo $id; ?>">click</a>

在你的情况下应该是

<td><a href="detail.php?id=<?php echo $rows["id"]; ?>">click</a></td>

你的while循环应该是这样的

while($rows = mysqli_fetch_array($SELECT)){ ?>
   <tr>
        <td><a href="paganame?id=<?php echo $rows["id"]; ?>">click</a></td>
        <td><?php echo $rows["name"]; ?></td>
        <td><?php echo $rows["xxx"]; ?></td>
        <td><?php echo $rows["xxx"]; ?></td>
        <td><?php echo $rows["xxx"]; ?></td>
        <td><?php echo $rows["xxx"]; ?></td>                                                                   <td>".$rows["xxx"]."</td>
    </tr>                                               
 <?php }