Question

我正在尝试搜索所有单词是否包含在长文本中。

SELECT *
FROM
  (SELECT substring_index(`sentence`, ' ', 1) AS first_keyword,
          substring_index(substring_index(`sentence`, ' ', 2), ' ', -1) second_keyword            
   FROM `keywords`) k
LEFT JOIN `article_table` a 
ON 
a.article_title LIKE concat('%', k.first_keyword, '%') AND 
a.article_title LIKE concat('%', k.second_keyword, '%')

http://sqlfiddle.com/#!9/ce6d67/5/0

这可以很好地工作，但是它需要很多SQL，但这些SQL不太可读，并且无法扩展，因为如果单词是3或4，我将无法获得相似的结果。

我完全陌生于编写存储过程。我找到了this example，但我想了解是否有更好的方法将句子拆分为单词。有什么建议吗？

DELIMITER $$

DROP PROCEDURE IF EXISTS `ContainsAllWords` $$
CREATE PROCEDURE `ContainsAnyWord`(_list MEDIUMTEXT)
BEGIN

DECLARE _next TEXT DEFAULT NULL;
DECLARE _nextlen INT DEFAULT NULL;
DECLARE _value TEXT DEFAULT NULL;

iterator:
LOOP
  -- exit the loop if the list seems empty or was null;
  -- this extra caution is necessary to avoid an endless loop in the proc.
  IF LENGTH(TRIM(_list)) = 0 OR _list IS NULL THEN
    LEAVE iterator;
  END IF;

  -- capture the next value from the list
  SET _next = SUBSTRING_INDEX(_list,',',1);

  -- save the length of the captured value; we will need to remove this
  -- many characters + 1 from the beginning of the string 
  -- before the next iteration
  SET _nextlen = LENGTH(_next);

  -- trim the value of leading and trailing spaces, in case of sloppy CSV strings
  SET _value = TRIM(_next);    

  -- rewrite the original string using the `INSERT()` string function,
  -- args are original string, start position, how many characters to remove, 
  -- and what to "insert" in their place (in this case, we "insert"
  -- an empty string, which removes _nextlen + 1 characters)
  SET _list = INSERT(_list,1,_nextlen + 1,'');
END LOOP;

END $$

DELIMITER ;

选择* 从keywords起左联接article_table a 上 a.article_title ContainsAllWords（k.sentence）

MySQL过程将字符串拆分为单词并搜索它们是否都包含在Text中

0 个答案: