mysql全文联合词

Question

我有一个 mysql 全文问题。参考：http://dev.mysql.com/doc/refman/5.0/en/fulltext-boolean.html

SELECT * FROM table WHERE (MATCH (aindex,bindex) AGAINST ('(+apple +juice)' IN BOOLEAN MODE))

这会返回 14 件商品

SELECT * FROM table WHERE (MATCH (aindex,bindex) AGAINST ('(+apple +pie)' IN BOOLEAN MODE))

这会返回 36 个项目

SELECT * FROM table WHERE (MATCH (aindex,bindex) AGAINST ('(+apple +pie)|(+apple +juice)' IN BOOLEAN MODE))

这返回 1910 个项目，显然，我的查询有问题。如何搜索包含(+apple +pie)和(+apple +juice)的单词？我认为应该是返回14+36=50items

在我的目的中，我需要一个联合搜索

(+apple +pie) (+apple +juice) (+apple +macintosh) (+apple +turnover) (+apple +strudel) (+pear +pie) (+pear +juice) (+pear +macintosh) (+pear +turnover) (+pear +strudel)

应该返回近 520 件商品。谢谢。

Answer 1

这应该是你想要的：

SELECT 
  * 
FROM 
  table 
WHERE 
  (MATCH (aindex,bindex) AGAINST ('+apple +(pie juice)' IN BOOLEAN MODE))

ETA：错过了我第一次阅读的梨子句。为了得到你列出的所有我认为这应该工作：

SELECT 
  * 
FROM 
  table 
WHERE 
  (MATCH (aindex,bindex) AGAINST ('+((apple +(pie juice macintosh turnover strudel)) (pear +(pie juice macintosh turnover strudel)))' IN BOOLEAN MODE));

这应该翻译成“必须包含苹果和馅饼、果汁、麦金塔、营业额或馅饼之一或梨和馅饼、果汁、麦金托什、营业额或馅饼之一。

如果这不起作用，我会这样做：

SELECT 
  * 
FROM 
  table 
WHERE 
  (MATCH (aindex,bindex) AGAINST ('+apple +(pie juice macintosh turnover strudel)' IN BOOLEAN MODE))
UNION
SELECT 
  * 
FROM 
  table 
WHERE 
  (MATCH (aindex,bindex) AGAINST ('+pear +(pie juice macintosh turnover strudel)' IN BOOLEAN MODE));