问题描述
以下是我已设置的3张桌子.通过索引零件表中零件名称中的所有单词,可以自动生成单词"和"word_part_mapping"表.
//'parts' table
+----------+--------------+
| part_num | part_name |
+----------+--------------+
| 10111 | front bumper |
| 10112 | rear bumper |
+----------+--------------+
//'words' table
+------+------------+
| id | word |
+------+------------+
| 1 | front |
| 2 | bumper |
| 3 | rear |
+------+------------+
//'word_part_mapping' association table
+---------+----------+
| word_id | part_num |
+---------+----------+
| 1 | 10111 |
| 2 | 10111 |
| 3 | 10112 |
| 1 | 10112 |
+---------+----------+
如果用户搜索前保险杠"并且希望查询返回包含用户搜索的所有单词的结果,该如何查询?
SELECT p.*
FROM words w
LEFT JOIN word_part_mapping wpm ON w.id=wpm.word_id
LEFT JOIN parts p ON wpm.part_num=p.part_num
WHERE w.word='front' AND w.word='bumper'
很显然,以上查询不起作用,因为单词不能同时等于"front"和"bumper".如果我执行或"操作,则此方法有效,但是我不希望这样做,因为它返回的结果太多(数据库中超过50,000个零件).
=============================================
终于可以使用了...
SELECT p.*
FROM words w
LEFT JOIN word_part_mapping wpm ON w.id=wpm.word_id
LEFT JOIN parts p ON wpm.part_num=p.part_num
WHERE w.word IN('front','bumper')
GROUP BY p.part_num
HAVING COUNT(DISTINCT w.word) = 2
其中2是用户正在搜索的字词数量
终于解决了...
SELECT p.*
FROM words w
LEFT JOIN word_part_mapping wpm ON w.id=wpm.word_id
LEFT JOIN parts p ON wpm.part_num=p.part_num
WHERE w.word IN('front','bumper')
GROUP BY p.part_num
HAVING COUNT(DISTINCT w.word) = 2
其中2是用户正在搜索的字词数量
Below are 3 tables I have setup. The 'words' and 'word_part_mapping' tables are generated automatically by indexing all of the words in the part's name from the parts table.
//'parts' table
+----------+--------------+
| part_num | part_name |
+----------+--------------+
| 10111 | front bumper |
| 10112 | rear bumper |
+----------+--------------+
//'words' table
+------+------------+
| id | word |
+------+------------+
| 1 | front |
| 2 | bumper |
| 3 | rear |
+------+------------+
//'word_part_mapping' association table
+---------+----------+
| word_id | part_num |
+---------+----------+
| 1 | 10111 |
| 2 | 10111 |
| 3 | 10112 |
| 1 | 10112 |
+---------+----------+
How can I query this if the user searches for "front bumper" and I want the query to return results containing ALL words that the user searched for?
SELECT p.*
FROM words w
LEFT JOIN word_part_mapping wpm ON w.id=wpm.word_id
LEFT JOIN parts p ON wpm.part_num=p.part_num
WHERE w.word='front' AND w.word='bumper'
Obviously, the above query does not work because the word cannot equal both 'front' and 'bumper'. This works if I do OR, however I do not want that because it returns too many results (50,000+ parts in database).
==============================================
EDIT: Got it working finally...
SELECT p.*
FROM words w
LEFT JOIN word_part_mapping wpm ON w.id=wpm.word_id
LEFT JOIN parts p ON wpm.part_num=p.part_num
WHERE w.word IN('front','bumper')
GROUP BY p.part_num
HAVING COUNT(DISTINCT w.word) = 2
where 2 is the number of terms the user is searching
Got it working finally...
SELECT p.*
FROM words w
LEFT JOIN word_part_mapping wpm ON w.id=wpm.word_id
LEFT JOIN parts p ON wpm.part_num=p.part_num
WHERE w.word IN('front','bumper')
GROUP BY p.part_num
HAVING COUNT(DISTINCT w.word) = 2
where 2 is the number of terms the user is searching
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