我有两张table。我想实现子行的显示/隐藏。我可以通过php得到相应的主子行。例如

ID      Field1   Field2
01        xxx      xxx
01        yyy      yyy
02        zzz      zzz
02        abc       abc
02        xyz      xyz

上表结构01 02 zzz zzz来自表1其他来自表2。每一行都有按钮。当我点击按钮时，相应的子行就会出现。问题是当我点击按钮 JS 仅返回第一行。我受够了。帮助我
我的php代码：（抱歉代码太多）

$connect = mysqli_connect('localhost', 'root', '', '');
$select1 =mysqli_query($connect,"SELECT * FROM ai");
while ($rownew=mysqli_fetch_array($select1)) 
{
?>
<input type='button' id="row_show_button<?php echo $rownew['idactionitems'];?>" value=<?php echo $rownew['idactionitems']?> onclick="ai_row_show_button('<?php echo $rownew['idactionitems'];?>');">
# some table datas here
</tr>
<?php 
 $new = mysqli_query($connect, "SELECT * FROM rowai WHERE idactionitemsFK =".$rownew['idactionitems']);
     while($newrow1=mysqli_fetch_array($new)) 
     {
?>
<tr style="display:" id="trhide<?php echo $newrow1['idactionitemsFK'];?>">
<td id="txt_name_edit1<?php echo $newrow1['idactionitemsFK'];?>"><?php echo $newrow1['idactionitemsFK'];?></td>
<td id="txt_name_edit2<?php echo $newrow1['idactionitemsFK'];?>"><?php echo $newrow1['row'];?></td>
</tr>
<?php
         }
}
?>

js代码：

function ai_row_show_button(idactionitems) {
    var trhide = document.getElementById("trhide"+idactionitems);
    //alert (trhide);
       if (trhide.style.display == "none") {
           <?php
$newsc = mysqli_query($connect, "SELECT * FROM rowai WHERE idactionitemsFK = idactionitemsFK ");
    #
     while($newrow1=mysqli_fetch_array($newsc)) 
     {         
           ?>
        trhide.style.display = "block";
     <?php } ?>
         } else {
        trhide.style.display = "none";
    }
}

我要寻找的解决方案是，当我单击第一个表生成的按钮时，按钮的值传递给 JS 功能和给予所有 <tr> 在第二个表中与id匹配。当前我只接收第一行。请帮帮我