Javascript 如何在Php中使用Select选项获取网格视图中的数据

我想在选择列表框项目时显示数据。以前在网格视图中，所有数据都将在那里。如果用户选择了一个选项，则只应显示选项数据。如何在Php中实现它。这是我的密码

 <td width="5%" align="right"><b>School<b></td><td width="1%"><b>:</b></td><td width="5%" align="left">
<select name="school_id" id="school_id" class="list_box">
<option value="">Select School</option>
<?php
$school_sql=mysql_query("select * from tbl_school");
while($school=mysql_fetch_array($school_sql))
{
?>
<option value="<?php echo $school['school_id'];?>" <?php if(isset($workshop_fetch['school_id']) && ($workshop_fetch['school_id']==$school['school_id'])) { echo "selected='selected'"; } ?> ><?php echo $school['school_name'];?></option>
<?php } ?>
</select>
 </td>
 </tr> 

  <tr class="table_txt2">
    <td width="7%">S.No</td>
    <td width="20%">School</td>
    <td width="10%">Payment Details</td>
    <td width="20%">Payment Date</td>
    <td width="13%">Amount</td>
    <td width="10%">Due Date</td>
    <td width="10%">Paid Date</td>
    <td width="10%">Is Paid</td>
  </tr>
  <?php
$slno=0;
$payment_sql=mysql_query("select * from tbl_school_payments");

while($result=mysql_fetch_array($payment_sql))
{
$slno++;

?>
<tr <?php if($slno%2==1) echo "class='table_txt3'"; else echo "class='table_txt4'"; ?> id="row_<?php echo $result['payment_id']; ?>">
<td ><?php echo $slno;?></td>
<td >
<?php
$school_sql=mysql_query("select * from tbl_school where school_id='$result[school_id]'");
$school=mysql_fetch_array($school_sql);
 echo ucwords($school['school_name']);
 ?>
</td>



//Here code for display data 
    <td ><?php echo $result['payment_details'];?></td>
    <td ><?php echo $result['payment_date'];?></td>
    <td ><?php echo $result['amount_paid'];?></td>
    <td ><?php echo $result['due_date'];?></td>
    <td ><?php echo $result['paid_date'];?></td>
    <td ><?php echo $result['is_paid'];?></td>

      <?php } ?>
      </table>

学校：
选择学校
学校：
选择学校
已经有一个解决方案了，请阅读此链接，还有更多的解决方案。但我想，如果你观察代码的第一行，有一个选择选项是学校。根据选择，只有学校必须显示。否则将显示学校总数。我想要这样。你可以用AJAX或者用
school\u id
向同一个URL发送GET参数，你能给我更多解释吗@user007I我会用GET方法写下答案。你能告诉我吗。如何使用ajaxYes实现这一点是的，我可以，但我不确定您在这里想要展示什么。如果从下拉列表中选择了一个学校id，那么所有select查询都使用该学校id仅显示这些记录，请您稍微弄清楚？
<td width="5%" align="right"><b>School<b></td><td width="1%"><b>:</b></td><td width="5%" align="left">
<select name="school_id" id="school_id" class="list_box">
<option value="">Select School</option>
<?php
$school_sql=mysql_query("select * from tbl_school");
while($school=mysql_fetch_array($school_sql))
{
?>
<option value="<?php echo $school['school_id'];?>" <?php if(isset($workshop_fetch['school_id']) && ($workshop_fetch['school_id']==$school['school_id'])) { echo "selected='selected'"; } ?> ><?php echo $school['school_name'];?></option>
<?php } ?>
</select>
 </td>
 </tr> 

  <tr class="table_txt2">
    <td width="7%">S.No</td>
    <td width="20%">School</td>
    <td width="10%">Payment Details</td>
    <td width="20%">Payment Date</td>
    <td width="13%">Amount</td>
    <td width="10%">Due Date</td>
    <td width="10%">Paid Date</td>
    <td width="10%">Is Paid</td>
  </tr>
  <?php
$slno=0;
$payment_sql=mysql_query("select * from tbl_school_payments");
while($result=mysql_fetch_array($payment_sql))
{
$slno++;

?>
<tr <?php if($slno%2==1) echo "class='table_txt3'"; else echo "class='table_txt4'"; ?> id="row_<?php echo $result['payment_id']; ?>">
<td ><?php echo $slno;?></td>
<td >
<?php
/*This IF block added
===================== */
if(isset($_GET['id']))
{

    $query = "select * from tbl_school where school_id=".mysql_real_escape_string($_GET['id']);
}
else
{
    $query = "select * from tbl_school where school_id='$result[school_id]'";
}
$school_sql=mysql_query($query);
$school=mysql_fetch_array($school_sql);
 echo ucwords($school['school_name']);
 ?>
</td>



//Here code for display data 
    <td ><?php echo $result['payment_details'];?></td>
    <td ><?php echo $result['payment_date'];?></td>
    <td ><?php echo $result['amount_paid'];?></td>
    <td ><?php echo $result['due_date'];?></td>
    <td ><?php echo $result['paid_date'];?></td>
    <td ><?php echo $result['is_paid'];?></td>

<?php } ?>
</table>

<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.0/jquery.min.js"></script>
<script type="text/javascript">

jQuery(document).ready(function($) {
      jQuery('#school_id').change(function(e){

          selected = jQuery('#school_id').val();
          window.location = window.location.pathname + '?id=' + selected;

      });
});