Question

我不知道如何添加表格结果的链接我想将<a href="#">#</a>更改为
<a href="photo_update.php?id=<?php echo $row['id']; ?>">edit</a>

提前感谢您的帮助。

foreach($results as $row) {
   $table  = '<tr>';
   $table .= '<td>';
   $table .= $row['filename'];
   $table .= '<a href="#">#</a>';
   $table .= '</td>';
   $table .= '<td class="col1 nw tar">';
   $table .= $row['size'];
   $table .= '</td>';
   $table .= '<td class="col1 nw tar">';
   $table .= $row['type'];
   $table .= '</td>';
   $table .= '<td class="col1 nw tar">&pound;';
   $table .= $row['caption'];
   $table .= '</td>';
   $table .= '</tr>';
}

Answer 1

在代码的第五行，而不是

$table .= '<a href="#">#</a>';

你应该写

$table .= '<a href="photo_update.php?id='.$row['id'].'">edit</a>';

Answer 2

foreach($results as $row) {

$table  = '<tr>';
$table .= '<td>';
$table .= $row['filename'];
$table .= "<a href='photo_update.php?id={$row['id']}'>edit</a>";
$table .= '</td>';
$table .= '<td class="col1 nw tar">';
$table .= $row['size'];
$table .= '</td>';
$table .= '<td class="col1 nw tar">';
$table .= $row['type'];
$table .= '</td>';
$table .= '<td class="col1 nw tar">&pound;';
$table .= $row['caption'];
$table .= '</td>';
$table .= '</tr>';

}

表和链接的结果

2 个答案: