以下是我的表格:POST数据阵列
<form method="POST" action="../controller/assignsubteacher.php">
<table class="table table-bordered table-striped table-hover table-condensed" id="coursedetail" >
<thead>
<tr>
<th>Sub Id</th>
<th>Sub Name</th>
<th>Teacher Name</th>
</tr>
</thead>
<tbody id="table_ajax">
</tbody>
<tfoot>
<tr>
<th>Sub Id</th>
<th>Sub Name</th>
<th>Teacher Name</th>
</tr>
</tfoot>
</table>
<div class="col-md-2">
<button type="submit" class="btn btn-primary btn-block btn-flat">Submit</button>
</div>
</form>
,并在形式表由以下响应填充:
while($row=mysqli_fetch_array($result))
{
$list='<select id="teacher" name="teacher'.$COUNT.'" class="form-control">
<option value = "UNKNOWN" selected="select">-SELECT-</option>';
$get_teacher="select Regno,Name from Student_Registration inner join Login on Regno=Uname where Id=2;";
$teacher_list = mysqli_query($con,$get_teacher);
while($row_Teacher=mysqli_fetch_array($teacher_list))
{
$list.='<option value="'.$row_Teacher['Regno'].'">'.$row_Teacher['Name'].'</options>';
}
$list.='</select>';
$Subject_ID=$row["SubId"].'<input type="hidden" name="SubId'.$COUNT.'" value="'.$row["SubId"].'">';
//$Subject_Name=$row["Subject_Name"].'<input type="hidden" name="SubName'.$COUNT.'" value="'.$row["Subject_Name"].'">';
$Subject_Name=$row["Subject_Name"];
$tr.='<tr>
<td>'.$Subject_ID.'</td>
<td>'.$Subject_Name.'</td>
<td>'.$list.'</td>
</tr>';
$COUNT=$COUNT+1;
}
echo $tr;
我不能够使用发布的数据中插入到数据库。有什么办法我可以发送数据作为数组并检索它。 下面是AJAX来填充表体:
xhr.onreadystatechange = function()
{
if (this.readyState == 4 && this.status == 200)
{
console.log(xhr.responseText);
Table.innerHTML=xhr.responseText;
}
};
我想使用的foreach插入在POST控制器的数据,但不知道如何实现这一目标。 任何帮助,将不胜感激。
什么是你得到的错误 –
任何解决方案? – Prakash