我有這個疑問:副按鈕值在單獨的表
$result = mysqli_query($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>";
}
這爲我提供了對每個星期的ID單獨的表。不過,我正在尋找按鈕來顯示下面的每個表與上面的結果相關聯。是否可以添加WEEK_ID值的按鈕。目前,如果我添加一個按鈕,頂部和底部與價值:
<input type='image' name='submit' src="image/button.jpg" value=" . $row['WEEK_ID'] . ">
它不會顯示爲頂級表進行正確的ID,它說明不了什麼了底部。我明白這是爲什麼,但無論如何,我可以將這個按鈕關聯在桌子下面嗎?
顯示所有非工作代碼,我們可以看到您在哪裏設置按鈕。 – TiMESPLiNTER
我在之後的按鈕代碼都在上面放置了按鈕。我只需要從上面一週的WEEK_ID中拿出自己的價值,但它並沒有將它與自身聯繫在一起。 – andy