只需将此添加到您的函数中。php。

Note: This filter works at the time of saving/updating the post.

add_filter( \'wp_insert_post_data\' , \'filter_post_data\' , \'99\', 2 );

function filter_post_data( $data , $postarr ) {

    $content = $data[\'post_content\'];

    $content = preg_replace(\'#<p.*?>(.*?)</p>#i\', \'<p>\\1</p>\', $content);
    $content = preg_replace(\'#<span.*?>(.*?)</span>#i\', \'<span>\\1</span>\', $content);
    $content = preg_replace(\'#<ol.*?>(.*?)</ol>#i\', \'<ol>\\1</ol>\', $content);
    $content = preg_replace(\'#<ul.*?>(.*?)</ul>#i\', \'<ul>\\1</ul>\', $content);
    $content = preg_replace(\'#<li.*?>(.*?)</li>#i\', \'<li>\\1</li>\', $content);

    $data[\'post_content\'] = $content;

    return $data;
}

Note: This filter works at the time when function the_content() is executed.

add_filter( \'the_content\', \'the_content_filter\', 20 );

function the_content_filter( $content ) {
    $content = preg_replace(\'#<p.*?>(.*?)</p>#i\', \'<p>\\1</p>\', $content);
    $content = preg_replace(\'#<span.*?>(.*?)</span>#i\', \'<span>\\1</span>\', $content);
    $content = preg_replace(\'#<ol.*?>(.*?)</ol>#i\', \'<ol>\\1</ol>\', $content);
    $content = preg_replace(\'#<ul.*?>(.*?)</ul>#i\', \'<ul>\\1</ul>\', $content);
    $content = preg_replace(\'#<li.*?>(.*?)</li>#i\', \'<li>\\1</li>\', $content);
    return $content;
}

我尝试了上面的保存/更新方法，但对我无效，所以我选择了另一种方法。我导出了整个wp\\u posts表，在Sublime中打开它并替换了一个regex。我用过style="*.*?" 找到所有的箱子，用空箱子代替。然后删除旧表的内容并导入新表。

如果有人尝试这种方法，请确保你有一个清晰的备份，以防wp\\u post表中有其他一些帖子类型，事情变得有点混乱。

我会查看content\\u save\\u pre filter，然后可能会应用一些奇特的正则表达式。