Php 如果发生验证错误，单选按钮值在编辑表单中不保留

我正在编辑表单中使用单选按钮选项。在编辑表单单选按钮选中依赖于数据库值。为此，我使用

<input type="radio"  name="active" value="active" <?php echo ($row['service']=='active')?'checked':'' ?>> Active
<input type="radio"  name="active" value="deactive" <?php echo ($row['service']=='deactive')?'checked':'' ?>> Deactive

>停用

工作正常。但在编辑表单中，如果有人更改此值并提交表单，但他保留了其他字段的任何验证错误，则单选按钮再次重置为旧值。我如何保留此值

例如，在用户编辑表单中，从数据库值中选择“停用”单选按钮，然后用户将此按钮更改为“启用”，并在“手机号码”字段中编辑，然后提交表单。如果“手机号码”字段中存在任何验证错误，则单选按钮也选择了“禁用”。如何保持它们不变

我的表格代码是

<?php include_once 'header.php';
$ra=$_SESSION['ra']
$sql="SELECT * FROM emitra_basic where uid='$ra'";
$result = $conn->query($sql);
$row=mysqli_fetch_array($result);
if(isset($_POST['update'])) 
     {$phone = $_POST['cno'];
     //validion example 
       if(strlen($phone)!=10) {
       $flag=1;
       $phoneErr = "Not a valid phone number";}.......like this other validaiton
   //then insertion..
    my form code are......
   <form id="basic" method="post" name="basic"  >
   <p class="contact"><label for="RU">service</label></p>               
    <input type="radio" id="active" name="active" value="active" <?php echo ($row['service']=='active')?'checked':'' ?>/> Active
    <input type="radio" id="active" name="active" value="deactive" <?php echo ($row['service']=='deactive')?'checked':'' ?>/> Deactive
    <p class="contact"><label for="contct No">Contact No</label></p> 
    <input id="cno" name="cno" placeholder="Contact No" value ="<?php  if(isset($phone)){ echo $phone; }elseif($phoneErr=="") {echo $row['contact'];} ?>" type="text"> 

<input type="radio" name="active" value="active" <?php if (isset($active) && $active=="active") echo "checked"; elseif($row['service']=='active') echo "checked"; ?> > Active
            <input type="radio"  name="active" value="deactive" <?php if (isset($active) && $active=="deactive") echo "checked"; elseif($row['service']=='deactive') echo "checked"; ?> > Deactive

<?php if (($_POST['active']) == 'active') {echo 'checked="checked"';} ?>