我正在制作一個頁面，該頁面將獲取我的所有數據庫值。我希望頁面從數據庫自動獲取Driver_id和Vehicle_id的值，用戶需要知道自己的ID和密鑰是什么。但是我被困在這里。我正在使用的工具是phpMyAdmin。下面是我的表格代碼：<!doctype html><html><style><table> <th>Vehicle ID</th> <th>Vehicle Model</th> <th>Vehicle Color</th> <th>Plate Number</th> <th>Seats</th> <th>Driver ID</th> <th> </th><?php $link=mysqli_connect("localhost","root","","jomsewa"); mysqli_select_db($link,"jomsewa") or die(mysqli_error($link)); $select = "SELECT * FROM vehicle"; $row = mysqli_query($link,$select); while ($array = mysqli_fetch_array($row)){ echo "<tr><td>".$array['Vehicle_id']."</td> <td>".$array['Vehicle_model']."</td> <td>".$array['Vehicle_color']."</td> <td>".$array['Vehicle_model']."</td> <td>".$array['Vehicle_seats']."</td> <td>".$array['Driver_id']."</td> <td><a href='Dmaintenance.php?Driverid=".$array['Driver_id']."'>Select</a></td>"."</tr>"; } mysqli_close($link);?></table></body></html>鏈接鏈接到Dmaintenance.php：<?php$link=mysqli_connect("localhost","root","","jomsewa"); if (!$link) { echo "Failed to connect to database: " . mysqli_connect_error(); }mysqli_select_db($link,"jomsewa") or die(mysqli_error($link));?><h3>Please update your maintenance details in the form below.</h3><form action="maintenance.php" method="post"><fieldset> <legend>Vehicle Maintenance Information:</legend> <table cellpadding="10"> <tr> <td> <?php if(isset($GET['Driver_id'])) { $txt = $GET['Driver_id']; while($row = mysqli_fetch_array($result)) { echo "<td>".$row['Vehicle_id']."</td>"; echo "<td>".$row['Driver_id']."</td>"; } }?></td> </tr>我想要的是，當單擊下一頁上的一個特定行鏈接時，它必須自動顯示我選擇的行內容。
2 回答
瀟湘沐
TA貢獻1816條經驗 獲得超6個贊
使用$_GET['Driverid]代替$_GET['Driver_id]
沒有SQL查詢Dmaintenance.php可獲取基于的行Driverid。應該有
$query = "SELECT * FROM vehicle WHERE Vehicle_id=".$_GET['Driverid'];
$row = mysqli_query($link,$query);
while ($array = mysqli_fetch_array($row)){
print_r($array);
}
例如
<a href="Dmaintenance.php?Driverid=123">Click Here</a>
并且僅在中使用following Dmaintenance.php，您將看到參數值
if(isset($_GET['Driverid'])){
echo $_GET['Driverid'];
}
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