Question

我有4张桌子。用户，文章，Langs，LangList

文章表： ...，uid（Users.id），lang（Langs.id），...

LangList表： id，uid（Users.id），langid（Langs.id）

我想为指定用户选择所有这些文章（例如Users.id = 66），这些文章位于他选择的语言列表中。

如果我这样做：

function FindActiveLangs($lang){
    $sql ="SELECT langid FROM LangList WHERE `uid`= '66' AND `active`='1';";
    $rsd = mysql_query($sql);
    $lang_string = " ( ".$lang."lang = '0' ";
    while($row = mysql_fetch_array($rsd)){
        $lang_string .= " OR ".$lang."lang = '".$row['langid']."' ";
    }
    $lang_string .= " ) ";
    return $lang_string;
} //end of FindActiveLangs


$tables = " Articles ";
$lang_string = FindActiveLangs("Articles.");
$where1 = "WHERE ".$lang_string." ";
$sql = "select distinct Articles.id, title, descr, thumb, relDate, 
            Articles.alias, Articles.lang, author, source  
        from ".$tables.$where1." and Users";

我在0.01-0.03秒内取得了结果

第二种方式，我认为清洁代码更好：

$sql = "select distinct Articles.id, title, descr, thumb, relDate, 
            Articles.alias, Articles.lang, author, source  
        from Articles 
        where lang IN 
            (SELECT langid FROM LangList WHERE uid= '66' AND active='1')";

我在1.1-1.3秒内取得了结果

有没有办法执行此查询，第二种方式具有更好的性能？ “IN”杀死速度

提前致谢...

Answer 1

查看EXISTS而不是IN。表现要好得多。

select distinct Articles.id, title, descr, thumb, relDate, Articles.alias, Articles.lang, author, source  
from Articles, TagsList 
where
    Articles.lang = 0 OR 
    EXISTS 
          (SELECT *
           FROM LangList 
           WHERE uid= '66' AND active='1' 
           AND langid = Articles.lang)

编辑：第二次尝试（检查表现是否更好）：

SELECT Articles.*
   FROM Articles 
    INNER JOIN LangList
        ON langid = Articles.lang
   WHERE uid= '66' AND active='1' 

UNION

SELECT Articles.*
FROM Articles
WHERE 
Articles.lang=0
AND
NOT EXISTS
(
    SELECT *
           FROM LangList 
           WHERE uid= '66' AND active='1' 
           AND langid = Articles.lang
)

在选择子查询中选择

1 个答案: