如何获得WordPress的language_attributes函数以返回有效的XHTML 1.1? [英] How can I get the WordPress language_attributes function to return valid XHTML 1.1?
问题描述
我有一个包含以下元素的WordPress模板:
I have a WordPress template that contains the following element:
<html xmlns="http://www.w3.org/1999/xhtml" <?php language_attributes('xhtml'); ?>>
这将返回:
<html xmlns="http://www.w3.org/1999/xhtml" dir="ltr" lang="en-US" xml:lang="en-US">
不幸的是，"lang"属性是无效的XHTML 1.1-客户端希望进行此级别的验证.
Unfortunately the "lang" attribute is invalid XHTML 1.1 - and the client would like this level of validation.
WordPress的general-template.php文件包含以下代码:
WordPress' general-template.php file contains the following code:
if ( get_option('html_type') == 'text/html' || $doctype == 'html' )
$attributes[] = "lang=\"$lang\"";
$doctype是传递给它的参数(在本例中为'xhtml'). get_option是否应该返回'text/html'以外的值?如果是这样，我应该在WordPress中设置什么来实现这一目标-如果有的话?
$doctype is the parameter passed to it (in this case 'xhtml'). Should get_option be returning a value other than 'text/html'? If so, what should I be setting in WordPress to achieve this - if anything?
我也尝试过使用preg_replace去掉"lang"属性，但是这似乎不能匹配文本.如果我手动输入文本，它将匹配！可能是language_attributes返回的字符串存在编码问题?
I've also tried using preg_replace to take out the "lang" attribute, but this didn't seem to be able to match the text. If I enter the text manually, it matches! Possibly an encoding issue with the string being returned by language_attributes?
推荐答案
我解决了这个问题.有一个"language_attributes"过滤器，所以我写了一个插件，它与此挂钩并执行了一个简单的preg_replace.在这里执行替换时，替换就可以了，这是一种很好的处理方式.
I solved this. There's a "language_attributes" filter, so I wrote a plugin that hooks into that and does a simple preg_replace. The replace worked when performed here, and it's a pretty neat way to handle it.
编辑
根据要求，这是我使用的代码:
As requested, here's the code I used:
<?php
/*
Plugin Name: Create Valid XHTML 1.1
Plugin URI: http://www.mycompany.com/create_valid_xhtml_1_1
Description: Removes deprecated "lang" attribute from (X)HTML header.
Author: dommer
Version: 1.0.0
Author URI: http://www.mycompany.com
*/
function create_valid_xhtml_1_1($language_attributes)
{
return preg_replace('/ lang=\"[a-z]+\-[A-Z]+\"/', '', $language_attributes);
}
add_filter('language_attributes', 'create_valid_xhtml_1_1');
?>
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