我有这个查询：
$result = mysqli_query($con,"SELECT * FROM b_tasks_report WHERE TASK_ID=$taskid ORDER BY WEEK_ID");
$current_week_id = -1;
while($row = mysqli_fetch_array($result))
{
if($current_week_id != $row['WEEK_ID'])
{
if($current_week_id != - 1)
{
echo "</table>";
}
echo "<table>";
echo "<tr class='no-border'><td class='no-border'><div class='task-detail-title'>Week Number: " . $row['WEEK_ID'] . "</div></td></tr>";
echo "<tr>";
echo "<th width='100'>Day</th>";
echo "<th width='75'>Start</th>";
echo "<th width='75'>End</th>";
echo "<th width='100'>Billable Hours</th>";
echo "<th width='100'>Non Billable Hours</th>";
echo "</tr>";
$current_week_id = $row['WEEK_ID'];
}
echo "<tr>";
echo "<td class='tdclass'>" . $row['DAY'] . "</td>";
echo "<td class='tdclass'>" . $row['START'] . "</td>";
echo "<td class='tdclass'>" . $row['END'] . "</td>";
echo "<td class='tdclass'>" . $row['BILLABLE_HOURS'] . "</td>";
echo "<td class='tdclass'>" . $row['NON_BILLABLE_HOURS'] . "</td>";
echo "</tr>";
}
if($current_week_id != - 1)
{
echo "</table>";
}
这为我提供了每周 ID 的单独表。 但是，我希望在每个表格下方显示与上述结果相关的按钮。 是否可以添加值为 WEEK_ID 的按钮。 目前，如果我在顶部和底部添加一个带有值的按钮：
<input type='image' name='submit' src="image/button.jpg" value=" . $row['WEEK_ID'] . ">
它不会为顶部表显示正确的 ID，也不会显示底部表的任何内容。 我确实明白为什么会这样，但是无论如何我可以在桌子下面关联这个按钮吗？
在
PHP 标签之外使用这个：
<input type="image" name="submit" src="image/button.jpg" value="<?php echo $row['WEEK_ID'] ?>">
在 PHP 标签中使用它。
echo '<input type="image" name="submit" src="image/button.jpg" value="' . $row['WEEK_ID'] . '">';