使用組中的所有關鍵字匹配MySQL

我有一個包含關鍵字（ID，ean，關鍵字）的表和包含產品詳細信息的另一個表。我希望搜索能夠返回所有關鍵字至少匹配一次的EAN，但是我得到的最接近的是以下內容，但是例如返回的匹配項中有第一項。舉個例子，假設我有一個名爲'通用耳機 - iPhone，iPad，iPod'的產品，並且我搜索'gen％''hea％''ip％'它會作爲匹配返回，但是它也會匹配'蘋果耳機 - iPhone，iPad，iPod'由於3個IP字，這是不希望的。使用組中的所有關鍵字匹配MySQL

SQL Fiddle 我想EAN 1只匹配，所以比賽需要至少1每個術語。

任何幫助將不勝感激。

 
SELECT  Count(keywords.ean) AS cc, 
      products.* 
FROM  keywords 
INNER JOIN products 
ON   products.ean = keywords.ean 
WHERE  ( 
         keyword LIKE 'gen%' 
           || keyword like 'ip%' 
           || keyword LIKE 'hea%') 
GROUP BY (keywords.ean) 
HAVING  cc>=3 
ORDER BY `products`.`ean` ASC

更新：這會得到所需的結果，但必須有更有效的方法來做到這一點。

 
SELECT products.* 
FROM products 
INNER JOIN (SELECT ean, count(*) as tc1 
        FROM keywords 
        WHERE (keyword like 'gen%') 
        GROUP BY ean 
        HAVING tc1 > 0) as t1 ON t1.ean = products.ean 
INNER JOIN (SELECT ean, count(*) as tc2 
        FROM keywords 
        WHERE (keyword like 'ip%') 
        GROUP BY ean 
        HAVING tc2 > 0) as t2 ON t2.ean = products.ean 
INNER JOIN (SELECT ean, count(*) as tc3 
        FROM keywords 
        WHERE (keyword like 'hea%') 
        GROUP BY ean 
        HAVING tc3 > 0 ) as t3 ON t3.ean = products.ean 
ORDER BY products.ean

來源

2017-06-04 dt192

見https://meta.stackoverflow.com/questions/333952/why-should-i -provide-an-mcve-for-what-seem-to-the-very-simple-sql-query – Strawberry

好吧，我已經用SQL小提琴更新了這篇文章。 – dt192

我認爲GregHNZ已經差不多了。只需將COUNT（*）位更改爲COUNT（DISTINCT關鍵字）= 3' – Strawberry

也許你更多的東西像這樣的後...

SELECT p.ean 
    , p.description 
    FROM products p 
    JOIN keywords k 
    ON k.ean = p.ean 
WHERE k.keyword LIKE 'iP%' 
    OR k.keyword LIKE 'hea%' 
    OR k.keyword LIKE 'gen%' 
GROUP 
    BY p.ean 
HAVING COUNT(DISTINCT CASE WHEN k.keyword LIKE 'iP%' THEN 'iP' 
          WHEN k.keyword LIKE 'hea%' THEN 'hea' 
          WHEN k.keyword LIKE 'gen%' THEN 'gen' 
          ELSE keyword END) = 3;

http://sqlfiddle.com/#!9/270f9/25

來源

2017-06-04 09:18:43 Strawberry

這就是我在PostgreSQL中的做法。 MySQL可能有一個稍微不同的語法。

SELECT kc.cc AS cc, 
     products.* 
FROM products 
INNER JOIN ( SELECT ean, count(*) AS cc 
        FROM keywords 
        WHERE (keyword like 'ip%' 
          OR keyword like 'ai%' 
          OR keyword like 'bei%') 
        GROUP BY ean 
        HAVING count(*) >= 3) AS kc 
     ON kc.ean = products.ean 
ORDER BY Products.ean;

來源

2017-06-04 01:38:09 GregHNZ

使用組中的所有關鍵字匹配MySQL

回答

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