優化多語言支持功能

Question

我爲自己的一個項目創建了自己的多語言支持。我有3個不同的數據庫表：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`) 
) 

當我想提取特定字符串的網站，我將使用一個名爲language()功能（如下圖所示）。下面代碼中的$user獲取有關已登錄用戶的信息，例如用戶選擇的語言。 （menu_about只是一個變量的例子）。

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; 
      } 

     } 

    } 

} 

問題是，網站加載非常緩慢，所有在language()內的東西。特別是當我將列出數據庫中的很多項目（例如一個循環）。加載速度從2秒到10秒不等。

是否有另一種更好，更快捷的方式來獲得所選的語言字符串？

UPDATE我的加載速度從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']; 

} 

我正在考慮使用文件（sv.php，en.php等）而不是數據庫。但是，那麼用戶不能從瀏覽器翻譯網站（如果我不犧牲自己幾個小時的血液，汗水和眼淚）。

Answer 1

首先分析構建Xhprof來驗證PHP性能。對於SQL調用Google Chrome有一個內置的Web檢查器，可以讓您知道每個組件的使用時間（Firebug for Firefox）。

如果您發佈日誌，它可以大大提高相關領域的改進。