0

我试图从两列中减去查询结果。

桌子：

 id |   word1   | lang1 |   word2   | lang2 |
----+-----------+-------+-----------+-------+
  1 | car       |   1   | car       |  15   |
  2 | table     |   1   | table     |  15   |
  3 | Chair     |   1   | cahair    |  13   |
  4 | CDplayer  |  15   | CDplayer  |   1   |
  5 | car       |   1   | car       |  13   |

我想得到语言 1 中 word1 中所有尚未翻译成语言 12 的单词。所以在这种情况下，它将是主席

表中有 300 万行，运行以下查询需要 1 分钟：

SELECT DISTINCT word1 
FROM `translations` 
WHERE lang1 = 1 
  AND lang2 != 15 
  AND NOT IN (SELECT word1 FROM `translations` WHERE lang2 == 15) 
LIMIT 10

分别对两行进行选择非常快 0.006 秒，然后我可以array_diff()在 PHP 中使用将它们彼此相减，但可能有一种更简单的方法可以直接在 MySQL 中进行。

4

4 回答 4

2 
SELECT 
    origin.word1
FROM 
    ( SELECT DISTINCT word1
      FROM tableX 
      WHERE lang1 = 1
    ) AS origin
WHERE 
    NOT EXISTS
      ( SELECT *
        FROM tableX AS trans
        WHERE trans.lang1 =  1
          AND trans.lang2 = 15
          AND trans.word1 = origin.word1
      ) ;

在运行这些查询之前，我会添加一个索引(lang1, word1)和一个索引。(lang1, lang2, word1)

您也可以尝试这种变化（并检查两个解释计划）：

SELECT DISTINCT
      word1
FROM 
      tableX AS origin
WHERE 
      lang1 = 1
  AND 
      NOT EXISTS
      ( SELECT *
        FROM tableX AS trans
        WHERE trans.lang1 =  1
          AND trans.lang2 = 15
          AND trans.word1 = origin.word1
      ) ;
于 2013-01-10T20:23:34.753 回答
0

我不确定我是否完全正确地理解了这个问题。聚合是否有效？

select word1
from translations t
group by word1
having max(lang1 = 1) = 1 and
       max(lang2 = 15) = 0
于 2013-01-11T00:23:50.917 回答
0

这应该很简单

SELECT
    m.id ,
    l.*
FROM mytable m
INNER JOIN (
            SELECT
                * 
            FROM mytable
            WHERE lang2 != 15
            GROUP BY id
) as l ON l.id = m.id
WHERE l.lang1 = 1
GROUP BY m.id
于 2013-01-10T19:58:42.120 回答
0 
SELECT DISTINCT NonTranslated.word1
from
(SELECT DISTINCT word1 FROM `translations` WHERE lang1 = 1 AND lang2 != 15)NonTranslated
left join
(SELECT DISTINCT word1 FROM `translations` WHERE lang1 = 1 AND lang2 = 15)Translated
on NonTranslated.word1 = Translated.word1
where Translated.word1 is NULL;

让我知道这是什么解释。我认为它可能比选择子查询更快。

PS：假设：即使这个词已经被翻译过一次，它也不会被包含在列表中。

于 2013-01-10T19:59:27.693 回答