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Php 将按钮值与单独表中的查询结果关联_Php_Sql - Fatal编程技术网

Php 将按钮值与单独表中的查询结果关联

Php 将按钮值与单独表中的查询结果关联,php,sql,Php,Sql,我有一个疑问: $result = mysqli_query($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)

我有一个疑问:

$result = mysqli_query($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>";
   }
$result=mysqli\u query($con,“从b\u tasks\u报告中选择*,其中TASK\u ID=$taskid ORDER BY WEEK\u ID”);
$current\u week\u id=-1;
while($row=mysqli\u fetch\u数组($result))
{
如果($current\u week\u id!=$row['week\u id']))
{
如果($current\u week\u id!=-1)
{      
回声“;
}
回声“;
回显“周编号:”.$row[“周ID]”;
回声“;
呼应“日”;
回应“开始”;
呼应“结束”;
回声“计费小时数”;
回声“非计费小时数”;
回声“;
$current_week_id=$row['week_id'];
}
回声“;
回显“$row['DAY']”;
回显“$row['START']”;
回显“$row['END']”;
回显“$row['bilable_HOURS'”;
回显“$行[“非计费小时数]”;
回声“;
}
如果($current\u week\u id!=-1)
{      
回声“;
}
这为每个星期ID提供了单独的表格。但是,我希望在每个表格下方显示与上述结果相关的按钮。是否可以添加一个值为WEEK_ID的按钮。当前,如果我在顶部和底部添加一个值为的按钮:

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


它没有为最上面的表显示正确的ID,也没有为最下面的表显示任何内容。我确实理解为什么会这样,但我是否可以在表下面关联此按钮?

在PHP标记之外使用此按钮:

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

显示整个非工作代码,我们可以看到按钮的确切设置位置。我将按钮放在上面两个按钮代码之后。我只需要从上面一周的WEEK_ID中获取值,但它没有将自身与之关联。很抱歉,它已经在PHP标记中。问题是它无法识别与上面表关联的WEEK\u ID。如果您想关联它,请给表一个
ID=“$row['WEEK\u ID']”
。再说一次,如果你告诉我们你想完成什么,答案可能会更直接。当你点击按钮时会发生什么?
echo '<input type="image" name="submit" src="image/button.jpg" value="' . $row['WEEK_ID'] . '">';