php - 计算匹配单词的数量

Question

我有两个表，数据填充在这个 sqlFiddle 中

现在，我有一个如下所示的查询，当我搜索“George Tabuki Street Fighter Miley Cyrus”时，我有 php explode 搜索字符串并动态构建查询添加 + CASE WHEN ... END

SELECT id,word,LEFT(description,100)as description, 
             IFNULL((SELECT sum(vote)
                     FROM vote v
                     WHERE v.definition_id = d.id),0) as votecount,
         0
    + CASE WHEN LOCATE('George',CONCAT(word,description,`usage`))>0 THEN 1 ELSE 0 END
    + CASE WHEN LOCATE('Tabuki',CONCAT(word,description,`usage`))>0 THEN 1 ELSE 0 END
    + CASE WHEN LOCATE('Street',CONCAT(word,description,`usage`))>0 THEN 1 ELSE 0 END
    + CASE WHEN LOCATE('Fighter',CONCAT(word,description,`usage`))>0 THEN 1 ELSE 0 END
    + CASE WHEN LOCATE('Miley',CONCAT(word,description,`usage`))>0 THEN 1 ELSE 0 END
    + CASE WHEN LOCATE('Cyrus',CONCAT(word,description,`usage`))>0 THEN 1 ELSE 0 END
as `match`
FROM definition d
HAVING `match` > 0
ORDER BY `match` DESC,votecount DESC

上面的查询返回的正是我想要的。

问题:有没有更好的方法，或者mySQL中有没有返回匹配单词数量的函数？

更新:

我找到了一种更好的方法，虽然不是更好的方法，但它返回匹配术语出现的次数

SELECT id,word,LEFT(description,100)as description, 
             IFNULL((SELECT sum(vote)
                     FROM vote v
                     WHERE v.definition_id = d.id),0) as votecount,
         0
        + IFNULL(ROUND((LENGTH(CONCAT(word,description,`usage`,`by`)) - LENGTH(REPLACE(CONCAT(word,description,`usage`,`by`), 'George', '')))/LENGTH('George')),0)
        + IFNULL(ROUND((LENGTH(CONCAT(word,description,`usage`,`by`)) - LENGTH(REPLACE(CONCAT(word,description,`usage`,`by`), 'Tabuki', '')))/LENGTH('Tabuki')),0)
        + IFNULL(ROUND((LENGTH(CONCAT(word,description,`usage`,`by`)) - LENGTH(REPLACE(CONCAT(word,description,`usage`,`by`), 'Street', '')))/LENGTH('Street')),0)
        + IFNULL(ROUND((LENGTH(CONCAT(word,description,`usage`,`by`)) - LENGTH(REPLACE(CONCAT(word,description,`usage`,`by`), 'Fighter', '')))/LENGTH('Fighter')),0)
        + IFNULL(ROUND((LENGTH(CONCAT(word,description,`usage`,`by`)) - LENGTH(REPLACE(CONCAT(word,description,`usage`,`by`), 'Miley', '')))/LENGTH('Miley')),0)
        + IFNULL(ROUND((LENGTH(CONCAT(word,description,`usage`,`by`)) - LENGTH(REPLACE(CONCAT(word,description,`usage`,`by`), 'Cyrus', '')))/LENGTH('Cyrus')),0)
as `match`
FROM definition d
HAVING `match` > 0
ORDER BY `match` DESC,votecount DESC;

Answer 1

不确定这是否是更好的方法，但我会这样做:

SELECT d.id, d.word, LEFT(d.description, 100) description,
  COALESCE(sum(v.vote), 0) votecount,
    (CONCAT(word, description, `usage`) LIKE '%George%')
  + (CONCAT(word, description, `usage`) LIKE '%Tabuki%')
  + (CONCAT(word, description, `usage`) LIKE '%Street%')
  + (CONCAT(word, description, `usage`) LIKE '%Fighter%')
  + (CONCAT(word, description, `usage`) LIKE '%Miley%')
  + (CONCAT(word, description, `usage`) LIKE '%Cyrus%') `match`
FROM definition d
LEFT JOIN vote v ON v.definition_id = d.id
GROUP BY d.id
HAVING `match` > 0
ORDER BY `match` DESC, votecount DESC

如果字符串足够长，则重复连接可能比创建派生表花费更多时间(不太可能，但值得一试):

SELECT id, word, description, votecount,
    (fullDesc LIKE '%George%')
  + (fullDesc LIKE '%Tabuki%')
  + (fullDesc LIKE '%Street%')
  + (fullDesc LIKE '%Fighter%')
  + (fullDesc LIKE '%Miley%')
  + (fullDesc LIKE '%Cyrus%') `match`
FROM (
  SELECT d.id, d.word, LEFT(d.description, 100) description,
    COALESCE(sum(vote), 0) votecount, CONCAT(word, description, `usage`) fullDesc
  FROM definition d
  LEFT JOIN vote v ON v.definition_id = d.id
  GROUP BY d.id
) s
HAVING `match` > 0
ORDER BY `match` DESC, votecount DESC