Question

我需要MYSQL搜索查询才能从我的表中获取热门话题，下面是我需要的解释

+----+---------+-----------------------------+
| ID | ID_user | text                        | 
+----+---------+-----------------------------+
| 1  | bruno   | michael jackson is dead     |
| 2  | thomasi | michael j. moonwalk is dead |
| 3  | userts  | michael jackson lives       |
+----+---------+-----------------------------+

我想查询表中最重复的单词，限制前10名，结果可能是这样：

+-------+------------+
| count | word       |
+-------+------------+
| 3     | michael    |
| 2     | dead       |
| 2     | jackson    |
| 1     | j.         |
| 1     | lives      |
| 1     | moonwalk   |
+-------+------------+

但是我想只搜索重复10次以上的单词，在这种情况下会出现没有单词，但如果重复单词的标准是2，它将只显示'michael'和'dead'，但忽略'is'因为我不想要用较少的2个字符长度的单词，以及一个短语的单词，然后我需要apear这个：

+-------+-----------------+
| count | word            |
+-------+-----------------+
| 2     | michael jackson |
| 2     | dead            |
+-------+-----------------+

Answer 1

CREATE TEMPORARY TABLE counters (id INT);
-- insert into counters as much as you like (values here means "number of repeats"
INSERT INTO counters VALUES (1), (2), (3), (4), (5), (6), (7), (8), (9),(10),
                           (11),(12),(13),(14),(15),(16),(17),(18),(19),(20),
                           (21),(22),(23),(24),(25),(26),(27),(28),(29),(30);

  SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(texts.text,' ',counters.id),' ',-1) AS word,
         COUNT(counters.id) AS counter
    FROM texts
         INNER JOIN counters ON (LENGTH(text)>0 AND SUBSTRING_INDEX(SUBSTRING_INDEX(text,' ',counters.id),' ',-1) <> SUBSTRING_INDEX(SUBSTRING_INDEX(text,' ',counters.id-1),' ', -1))
   WHERE length(SUBSTRING_INDEX(SUBSTRING_INDEX(texts.text,' ',counters.id),' ',-1)) > 2
GROUP BY word
  HAVING COUNT(counters.id) > 1
ORDER BY counter desc;

但效率不高，不应该那么做

编辑：

  SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(texts.text,' ',counters.id),' ',-1) AS word,
         COUNT(counters.id) AS counter
    FROM texts
         INNER JOIN counters ON (LENGTH(text)>0 AND SUBSTRING_INDEX(SUBSTRING_INDEX(text,' ',counters.id),' ',-1) <> SUBSTRING_INDEX(SUBSTRING_INDEX(text,' ',counters.id-1),' ', -1))
   -- exclude words list
   WHERE SUBSTRING_INDEX(SUBSTRING_INDEX(texts.text,' ',counters.id),' ',-1) NOT IN ('is', 'of', 'this', 'to')
GROUP BY word
  HAVING COUNT(counters.id) > 1
ORDER BY counter desc;

搜索查询，如Twitter趋势主题

1 个答案: