优化多语言支持功能

Question

我为自己的一个项目创建了自己的多语言支持。我有3个不同的数据库表：languages,languages_variables和languages_words（如下所示）。

CREATE TABLE IF NOT EXISTS `languages` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `id_user` int(11) NOT NULL, 
    `data_name` text NOT NULL, 
    `data_currency` text NOT NULL, 
    `is_standard` tinyint(4) NOT NULL, 
    `data_note` text NOT NULL, 
    PRIMARY KEY (`id`), 
    UNIQUE KEY `id` (`id`) 
) 

CREATE TABLE IF NOT EXISTS `languages_variables` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `data_name` text NOT NULL, 
    PRIMARY KEY (`id`), 
    UNIQUE KEY `id` (`id`) 
) 

CREATE TABLE IF NOT EXISTS `languages_words` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `id_language` int(11) NOT NULL, 
    `id_variable` int(11) NOT NULL, 
    `data_name` text NOT NULL, 
    PRIMARY KEY (`id`), 
    UNIQUE KEY `id` (`id`) 
) 

当我想提取特定字符串的网站，我将使用一个名为language()功能（如下图所示）。下面代码中的$user获取有关已登录用户的信息，例如用户选择的语言。 （menu_about只是一个变量的例子）。

function language($variable) { 

    global $user; 


    # DATABAS (kontrollera) 
    $check_language = sql("SELECT COUNT(lv.data_name) 

          FROM languages AS l 
          JOIN languages_variables AS lv 

          JOIN languages_words AS lw 
          ON l.id = lw.id_language 

          WHERE lv.data_name = :variable 
          AND lw.id_language = :idlanguage 
          ", Array('idlanguage' => ($user['id_language'] == 0 ? '1' : (int)$user['id_language']), 
            'variable' => $variable), 'count'); 

    # DATABAS (räkna) 
    $count_words = sql("SELECT COUNT(lw.id) 
         FROM languages_words AS lw 
         JOIN languages_variables AS lv 
         ON lw.id_variable = lv.id 
         WHERE lw.id_language = :idlanguage 
         AND lv.data_name = :variable 
         ", Array('idlanguage' => ($user['id_language'] == 0 ? '1' : (int)$user['id_language']), 
           'variable' => $variable), 'count'); 



    # KONTROLL 
    if($check_language == 0 OR $count_words == 0) { 

     return $variable; 


    # KONTROLL 
    } else { 

     # DATABAS (hämta) 
     $get_languages = sql("SELECT *, 
            lv.data_name AS name_variable, 
            lw.data_name AS name_translation 

           FROM languages AS l 
           JOIN languages_variables AS lv 

           JOIN languages_words AS lw 
           ON lv.id = lw.id_variable 

           WHERE lv.data_name = :variable 
           AND lw.id_language = :idlanguage 
          ", Array('variable' => $variable, 
             'idlanguage' => ($user['id_language'] == 0 ? '1' : (int)$user['id_language']))); 


     # LOOP 
     foreach($get_languages AS $language) { 

      # KONTROLL 
      if($language['name_variable'] == $variable) { 
       return $language['name_translation']; 

      # KONTROLL 
      } else { 
       return $variable; 
      } 

     } 

    } 

} 

问题是，网站加载非常缓慢，所有在language()内的东西。特别是当我将列出数据库中的很多项目（例如一个循环）。加载速度从2秒到10秒不等。

是否有另一种更好，更快捷的方式来获得所选的语言字符串？

UPDATE我的加载速度从7-8秒降到3-4秒。但它仍然太慢。

function language($variable) { 

    # DATABAS (hämta) 
    $language = sql("SELECT lv.data_name AS name_variable, 
          lw.data_name AS name_word 

        FROM languages AS l 
        JOIN languages_variables AS lv 
        JOIN languages_words AS lw 

        ON lv.id = lw.id_variable 
        WHERE lv.data_name = :variable 
        AND l.id = :idlanguage 
        ", Array('variable' => $variable, 
          'idlanguage' => ($user['id_language'] == 0 ? '1' : (int)$user['id_language'])), 'fetch'); 


    return $language['name_word']; 

} 

我正在考虑使用文件（sv.php，en.php等）而不是数据库。但是，那么用户不能从浏览器翻译网站（如果我不牺牲自己几个小时的血液，汗水和眼泪）。

Answer 1

首先分析构建Xhprof来验证PHP性能。对于SQL调用Google Chrome有一个内置的Web检查器，可以让您知道每个组件的使用时间（Firebug for Firefox）。

如果您发布日志，它可以大大提高相关领域的改进。