如何以不同的样式显示php echo结果？

我有两个来自mysql数据库的php echo结果。他们的请求很常见，并被包装成一行：

$sql = "SELECT * FROM test WHERE id IN ('9','10')";

    <?php
        $conn = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
        if ($conn->connect_error)
        {
                echo "connection error" ;
        }
        $sql = "SELECT * FROM test WHERE id IN ('9','10')";
        $result = $conn->query($sql);
    ?>
    <?php
    $i = 0;
    if ($result->num_rows > 0) { 
            while($data = $result->fetch_assoc()){
                  if($i == 0){
                        echo "<h3>".$data ['FR']."<br></h3>"; 
                    }else{
                        echo "<p>".$data ['FR']."<br></p>"; 
                    }
                     $i++;
            }    
    } else {
        echo "---"; 
    }
    ?>

我只希望第一个结果以h3样式显示，第二个结果以正常p样式显示。使用下面的代码两个结果以h3样式显示。

    <?php
        $conn = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
        if ($conn->connect_error)
        {
                echo "connection error" ;
        }
        $sql = "SELECT * FROM test WHERE id IN ('9','10')";
        $result = $conn->query($sql);
    ?>
    <?php
    $i = 0;
    if ($result->num_rows > 0) { 
            while($data = $result->fetch_assoc()){
                  if($i == 0){
                        echo "<h3>".$data ['FR']."<br></h3>"; 
                    }else{
                        echo "<p>".$data ['FR']."<br></p>"; 
                    }
                     $i++;
            }    
    } else {
        echo "---"; 
    }
    ?>

我怎么修理它

    <?php
        $conn = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
        if ($conn->connect_error)
        {
                echo "connection error" ;
        }
        $sql = "SELECT * FROM test WHERE id IN ('9','10')";
        $result = $conn->query($sql);
    ?>
    <?php
    $i = 0;
    if ($result->num_rows > 0) { 
            while($data = $result->fetch_assoc()){
                  if($i == 0){
                        echo "<h3>".$data ['FR']."<br></h3>"; 
                    }else{
                        echo "<p>".$data ['FR']."<br></p>"; 
                    }
                     $i++;
            }    
    } else {
        echo "---"; 
    }
    ?>

先谢谢你

enter code here

<div class="one-third wow fadeIn">
<span class="circle"><span class="ico pig"></span></span>
<h3>

<?php
$conn = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
if ($conn->connect_error)
{
echo "connection error" ;
}

$sql = "SELECT * FROM test WHERE id IN ('9','10')";
$result = $conn->query($sql);

?>
<?php
if ($result->num_rows > 0) { while($data = $result->fetch_assoc())
{echo $data ['FR']."<br>"; }    } else {echo "---"; }?></h3>
<p>

    <?php
        $conn = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
        if ($conn->connect_error)
        {
                echo "connection error" ;
        }
        $sql = "SELECT * FROM test WHERE id IN ('9','10')";
        $result = $conn->query($sql);
    ?>
    <?php
    $i = 0;
    if ($result->num_rows > 0) { 
            while($data = $result->fetch_assoc()){
                  if($i == 0){
                        echo "<h3>".$data ['FR']."<br></h3>"; 
                    }else{
                        echo "<p>".$data ['FR']."<br></p>"; 
                    }
                     $i++;
            }    
    } else {
        echo "---"; 
    }
    ?>

在此处输入代码
试试这个


    <?php
        $conn = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
        if ($conn->connect_error)
        {
                echo "connection error" ;
        }
        $sql = "SELECT * FROM test WHERE id IN ('9','10')";
        $result = $conn->query($sql);
    ?>
    <?php
    $i = 0;
    if ($result->num_rows > 0) { 
            while($data = $result->fetch_assoc()){
                  if($i == 0){
                        echo "<h3>".$data ['FR']."<br></h3>"; 
                    }else{
                        echo "<p>".$data ['FR']."<br></p>"; 
                    }
                     $i++;
            }    
    } else {
        echo "---"; 
    }
    ?>

是的，它在工作，克兰蒂！你是头儿！谢谢您。非常地！