PHP + Joomla(Sourcerer插件)。任何明显的错误?

时间:2015-11-16 20:12:39

标签: php joomla

这是我的文章。 PHP代码包含在{source}标记中,而sourcerer在文章中神奇地运行PHP。我以前成功使用过Sourcerer;但现在却抛出了一些奇怪的错误。

{source}

<?php

if (isset($_POST['submit'])) {
    $connect = mysql_connect("host","user","pass"); 

    mysql_select_db("<redacted>", $connect);

    $term = $_POST['userid'];
    $getUserid = mysql_query("SELECT itemid, first, last FROM tenants where CONCAT(first, last) LIKE '%$term%'");
    $result = mysql_fetch_array($getUserId);
?>
{/source}

<table>
    <tr>
        <td align="center">itemID</td>
        <td align="center">First</td>
        <td align="left">Last</td>
    </tr>
    {source}
    <?php 
    while ($data = mysql_fetch_row($result)) {
    ?>
    {/source}
    <tr>
        <td align="center" valign="middle">
            {source} <?php echo $data['id']; ?> {/source}
        </td>
        <td align="center" valign="middle">
            {source} <?php echo $data['first']; ?>{/source}
        </td>
        <td align="left" valign="middle">
            {source} <?php echo $data['last']; ?>{/source}
        </td>
    </tr>
</table>
{source}
<?php
} 
}
?>
{/source}

Enter a Search Term: <input name="submit" type="submit" value="Register Now" style="width: 130px; height: 30px;" />

访问文章时的错误是:

[Mon Nov 16 15:03:41 2015] [warn] [client 74.84.255.79] mod_fcgid: stderr: PHP Parse error:  syntax error, unexpected end of file in /mnt/data/vhosts/<redacted>/plugins/system/sourcerer/helper.php(607) : runtime-created function on line 17, referer: http://host/
[Mon Nov 16 15:03:41 2015] [warn] [client 74.84.255.79] mod_fcgid: stderr: PHP Fatal error:  Function name must be a string in /mnt/data/vhosts/<redacted>/plugins/system/sourcerer/helper.php on line 612, referer: http://host/

在helper.php中,这是第607行:

$temp_PHP_func = create_function('&$src_vars, &$article, &$Itemid, &$mainframe, &$app, &$document, &$doc, &$database, &$db, &$user', $src_script);

这是612:

$src_new_vars = $temp_PHP_func($src_vars, $article, $Itemid, $mainframe, $app, $document, $doc, $database, $db, $user);

我会感激任何帮助。我在这里完全失去了。

0 个答案:

没有答案