使用echo创建amp-html代码时无法找到PHP语法错误

Question

我试图用PHP回应一些amp-html代码

<?php
echo "<section>";
echo "<h3>$title[$i]</h3>";
echo "<amp-video controls width="560" height="315" layout="responsive">";
echo "<source src="https: //example.com/videos/";
    echo $filename[$i];
    echo "_HD.mp4 type="video / mp4 / > ";
    echo " < sourcesrc = "https://example.com/videos/";
    echo $filename[$i];
    echo '"__HD.webm" type="video/webm"/>';
    echo "<div fallback><p>This browser does not support the video element.</p></div>";
    echo "</amp-video>";
    echo "</section>";
?>

我一直无法找到语法错​​误，我错过了什么？有没有比使用echo更简单的方法。

更新：我已经根据建议编辑了代码，但它仍然抛出一个空白页面：这是完整的代码：

<?php
        // Create connection
        $con=mysqli_connect("localhost","xx","xxxx","xxxx");

       // Check connection
    if (mysqli_connect_errno($con)) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    //Get number of rows

$sql="SELECT id,title,filename FROM videos";

    $result=mysqli_query($con, $sql);
    $i=1;
    while($row=mysqli_fetch_assoc($result))
    {
        $id[$i] = $row['id'];
        $title[$i] = $row['title'];
        $filename[$i] = $row['filename'];
        $i++;
    }
    // Loop through the results from the database

    for ($i = 1; $i <=count($id); $i++)
    {
?>

<section>
    <h3><?php echo $title[$i];?></h3>
    <amp-video controls width="560" height="315" layout="responsive">
        <source src="https: //example.com/videos/<?php echo $filename[$i];?>_HD.mp4" type="video/mp4" />
        <source src="https://example.com/videos/<?php echo $filename[$i];?>__HD.webm" type="video/webm" />
        <div fallback><p>This browser does not support the video element.</p></div>
    </amp-video>
</section>
}
</amp-accordion>

Answer 1

由于代码中的实际PHP输出很少，因此只需退出PHP即可输出HTML，这样可能更容易，更易读。

        // Create connection
        $con=mysqli_connect("localhost","xx","xxxx","xxxx");

       // Check connection
    if (mysqli_connect_errno($con)) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    //Get number of rows

$sql="SELECT id,title,filename FROM videos";

    $result=mysqli_query($con, $sql);
    while($row=mysqli_fetch_assoc($result)) {
?>

<section>
    <h3><?php echo $row['title']; ?></h3>
    <amp-video controls width="560" height="315" layout="responsive">
        <source src="https: //example.com/videos/<?php echo $row['filename']; ?>_HD.mp4" type="video/mp4" />
        <source src="https://example.com/videos/<?php echo $row['filename']; ?>__HD.webm" type="video/webm" />
        <div fallback><p>This browser does not support the video element.</p></div>
    </amp-video>
</section>
<?php
    }  // end of while
?>
</amp-accordion>
<?php

Answer 2

问题确实是使用您的语法。你应该尝试做的是当你使用html的回声时使用单引号，以及你在html中使用双引号添加的任何其他内容。

  <?php
echo '<section>';
echo '<h3>$title[$i]</h3>';
echo '<amp-video controls width="560" height="315" layout="responsive">';
echo '<source src="https://example.com/videos/'.$filename[$i].'_HD.mp4" type="video / mp4 / >"';
    echo '<source src="https://example.com/videos/'.$filename[$i].'__HD.webm" type="video/webm"/>"';
    echo '<div fallback><p>This browser does not support the video element.</p></div>';
    echo '</amp-video>';
    echo '</section>';
?>

通过这种方式，可以更容易地发现语法错误，因为页面将显示必须显示的内容或显示代码。