Question

我有一个名为“content”的MySQL表，其中包含（a.o.）字段“_date”和“text”，例如：

_date      text
---------------------------------------------------------
2011-02-18 I'm afraid my car won't start tomorrow
2011-02-18 I hope I'm going to pass my exams
2011-02-18 Exams coming up - I'm not afraid :P
2011-02-19 Not a single f was given this day
2011-02-20 I still hope I passed, but I'm afraid I didn't
2011-02-20 On my way to school :)

我正在寻找一个查询来计算每天使用“希望”和“害怕”字样的次数。换句话说，输出必须是：

_date      word   count
-----------------------
2011-02-18 hope   1
2011-02-18 afraid 2
2011-02-19 hope   0
2011-02-19 afraid 0
2011-02-20 hope   1
2011-02-20 afraid 1

有没有一种简单的方法可以做到这一点，或者我应该在每个学期写下不同的查询？我现在有这个，但我不知道该放什么而不是“？”

SELECT COUNT(?) FROM content WHERE text LIKE '%hope' GROUP BY _date

有人可以帮忙解决这个问题吗？

Answer 1

我认为最简单易行的方法是制作subquerys：

 Select 
    _date, 'hope' as word, 
    sum( case when `text` like '%hope%' then 1 else 0 end) as n
 from content
 group by _date
 UNION
 Select 
    _date, 'afraid' as word, 
    sum( case when `text` like '%afraid%' then 1 else 0 end) as n
 from content
 group by _date

这种方法效果不佳。如果您正在寻找性能，那么您应该在白天对子查询进行分组，此like条件也是性能杀手。如果您只是按批处理模式执行查询，那么这是一个解决方案。解释您的性能要求以获得准确的解决方案。

编辑以匹配最后的OP要求

Answer 2

您的查询几乎正确无误：

SELECT _date, 'hope' AS word, COUNT(*) as count
FROM content WHERE text LIKE '%hope%' GROUP BY _date

使用%hope%匹配任何地方的单词（不仅仅是在字符串的末尾）。 COUNT(*)应该做你想做的事。

要从单个查询中获取多个字词，请使用UNION ALL

另一种方法是动态创建一系列单词并将其用作连接中的第二个表：

SELECT _date, words.word, COUNT(*) as count
FROM (
   SELECT 'hope' AS word
   UNION
   SELECT 'afraid' AS word
) AS words
CROSS JOIN content
WHERE text LIKE CONCAT('%', words.word, '%')
GROUP BY _date, words.word

请注意，每个句子只会计算每个单词的一次出现次数。所以»我希望仍有希望«只会给你1，而不是2

要在没有匹配项时获取0，请再次将上一个结果与日期相结合：

SELECT content._date, COALESCE(result.word, 'no match'), COALESCE(result.count, 0)
FROM content
LEFT JOIN (
SELECT _date, words.word, COUNT(*) as count
FROM (
   SELECT 'hope' AS word
   UNION
   SELECT 'afraid' AS word
) AS words
CROSS JOIN content
WHERE text LIKE CONCAT('%', words.word, '%')
GROUP BY _date, words.word ) AS result
ON content._date = result._date

Answer 3

假设您想要计算所有单词并找到最常用的单词（而不是查找几个特定单词的计数），您可能需要尝试类似下面的存储过程（this blog post的字符串拆分补语）：

DROP PROCEDURE IF EXISTS wordsUsed;
DELIMITER //
CREATE PROCEDURE wordsUsed ()
BEGIN
    DROP TEMPORARY TABLE IF EXISTS wordTmp;
    CREATE TEMPORARY TABLE wordTmp (word VARCHAR(255));

    SET @wordCt  = 0;
    SET @tokenCt = 1;

    contentLoop: LOOP
        SET @stmt = 'INSERT INTO wordTmp SELECT REPLACE(SUBSTRING(SUBSTRING_INDEX(`text`, " ", ?),
                                LENGTH(SUBSTRING_INDEX(`text`, " ", ? -1)) + 1),
                                " ", "") word
                     FROM content
                     WHERE LENGTH(SUBSTRING_INDEX(`text`, " ", ?)) != LENGTH(`text`)';
        PREPARE cmd FROM @stmt;
        EXECUTE cmd USING @tokenCt, @tokenCt, @tokenCt;
        SELECT ROW_COUNT() INTO @wordCt;
        DEALLOCATE PREPARE cmd;
        IF (@wordCt = 0) THEN
            LEAVE contentLoop;
        ELSE
            SET @tokenCt = @tokenCt + 1;
        END IF;
    END LOOP;

    SELECT word, count(*) usageCount FROM wordTmp GROUP BY word ORDER BY usageCount DESC;
END //
DELIMITER ;

CALL wordsUsed();

您可能想要编写另一个查询（或过程）或添加一些嵌套的“REPLACE”语句，以进一步从生成的临时表中删除标点符号，但这应该是一个好的开始。

计算每天使用一个单词的次数

3 个答案: