Question

我有这个问题：

$result = mysqli_query($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的单独表格。但是我希望在每个表格下面显示与上面结果相关的按钮。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'] . ">

它没有显示顶部表格的正确ID，并且底部没有显示任何内容。我明白为什么会这样，但无论如何我可以将这个按钮关联在桌子下面吗？

Answer 1

在PHP标记之外使用它：

<input type="image" name="submit" src="image/button.jpg" value="<?php echo $row['WEEK_ID'] ?>">

在PHP标记内使用它。

echo '<input type="image" name="submit" src="image/button.jpg" value="' . $row['WEEK_ID'] . '">';

将按钮值与单独表中的查询结果相关联

1 个答案: