1

请帮助，我在 ?> 之后没有添加任何内容，我试图将代码放在 echo 中，但它不起作用请有人为我整理一个 JSFiddle 或为我指出正确的方向，我对 PHP 相当陌生，谢谢您的帮助

<?php
    $filepath = 'http://www.godsgypsychristianchurch.net/music.json';
    $content = file_get_contents($filepath);
    $json = json_decode($content, true);


foreach ($json['rows'] as $row)
    {
            if ($_GET['album'] == $row[doc]['album'])
            {
                        echo "<title>{$row[doc]['album']}</title>";
                        echo "<table align=\"center\" border=\"0\"><tr><td valign=\"top\" width=\"330\">";
                        echo "<img src=\"{$row['doc']['artwork']}\" alt=\"my image \" width=\"250\" /><br /><br />";
                        echo "<div class=\"albuminfo\" id=\"albuminfo\">";
                    print ' <a href=http://ggcc.tv/archivealbum.php?download=' . urlencode($row[doc]['album']) . ' id=DownloadAlbum><img src="http://ggcc.tv/musiclibrary/wp-content/plugins/zina/zina/themes/zinaEmbed/icons/download.gif" width="35">Download entire album.</a><p>';
                        echo "<font color=\"#fff\">Album: {$row[doc]['album']}</font><br />";
                        echo "<font color=\"#fff\">Church: {$row[doc]['church']}</font><br />";
                        echo "<font color=\"#fff\">Description: {$row[doc]['des']}</font><P><br /><P>";
                        echo "<a href=\"https://twitter.com/share\" class=\"twitter-share-button\" data-lang=\"en\">Tweet</a><br><br>";
                        print '<div id="like-button"></div>';
                        echo "<td valign=\"top\">";
                        echo "<div class=\"playlist\" id=\"playlist\">";
                        echo "<ol>";
                        $songCount = 0;
                        foreach ($row['doc']['tracks'] as $song) {
                            ++$songCount;

                            $songUrl = $row['doc']['baseurl'] . urldecode($song['url']);
                            echo "<li><a href=\"#\" data-src=\"{$songUrl}\">{$song['name']}</a><div id=\"download\"><a href=\"{$songUrl}\">Download</a></li>";
                        }
                        echo "</ol>";
                        echo "<br><div id=\"player\"><audio preload></audio></div>";
                        echo "</div>";
                        echo "<P>";
                        echo "<small>To download a single MP3 at a time:</br><b>Windows OS:</b> hold the ALT button on the keyboard and click the Download button<br><b>Mac OSX:</b> hold the OPTION button on the keyboard and click the Download button<P><BR><b>Controls:</b><br>Play/Pause = spacebar</br>Next track = Right arrow<br>Previous track = Left arrow";
                        echo '</tr></td></table>';
    }



            }



    exit;
?>
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4

3 回答 3

7

因为您正在使用exit，它会终止脚本。摆脱它，它将继续在下面输出 HTML。

http://php.net/manual/en/function.exit.php

于 2013-02-03T10:47:58.697 回答
1

省略出口。

退出将结束 PHP 处理。

于 2013-02-03T10:48:05.883 回答
1

随着 php exit 解析器停止，并将收集到的所有内容交还给 Web 服务器。只需删除退出；排。

于 2013-02-03T10:48:21.060 回答