Php ECHO MYSQL结果显示空白页

我知道这一定很简单，但我做不到

我需要将文本显示为DIV

像这样：

<div>
  <p><font color="#bdbdbd">TEXTO FROM MYSQL</font></p>
</div>

您可以在这里看到我的代码：loja.genesiseseries/depomentos/testemsql.html

$sql = "SELECT * FROM opinions ORDER BY id DESC LIMIT 15";
$resultado = mysql_query($sql);

while ($linha=mysql_fetch_array($resultado)) {
    $depoimento = $linha["depoimento"];
    $client = $linha["client"];
    echo "$depoimento";
    echo "$client";
}
?>

但它仍然出现在空白处我使用的是一个半专用服务器（VPS）

有什么想法吗

---

谢谢

将

loja.genesseries/depomentos/testemsql.html

重命名为

loja.genesseries/depomentos/testemsql.php

<div>
    <p>
        <font color="#bdbdbd">
            <?php
            $sql = "SELECT * FROM opinions ORDER BY id DESC LIMIT 15";
            $resultado = mysql_query($sql);
            while ($linha=mysql_fetch_array($resultado)) {
                $depoimento = $linha["depoimento"];
                $client = $linha["client"];
                echo "$depoimento";
                echo "$client";
            }
            ?>
        </font>
    </p>
</div>

Php和html是不同的…你的评论应该如何帮助我Deepu？阅读应该有帮助。可能重复感谢Ian Brindley！！但是有没有办法将此代码添加到HTML页面？@user2415711没有。该页面需要是一个php文件。但是，您可以使用mod_rewrite使其看起来像地址栏中的

.html

文件

<div>
    <p>
        <font color="#bdbdbd">
            <?php
            $sql = "SELECT * FROM opinions ORDER BY id DESC LIMIT 15";
            $resultado = mysql_query($sql);
            while ($linha=mysql_fetch_array($resultado)) {
                $depoimento = $linha["depoimento"];
                $client = $linha["client"];
                echo "$depoimento";
                echo "$client";
            }
            ?>
        </font>
    </p>
</div>