Question

SELECT 
  games.id AS id,
  games.ean AS ean,
  games.title AS title,
  games.developer AS developer,
  reviews.NoReviews AS NoReviews 
FROM
  games 
  INNER JOIN 
    (SELECT 
      ean,
      COUNT(*) AS NoReviews 
    FROM
      reviews 
    GROUP BY ean) reviews 
    ON games.ean = reviews.ean 
WHERE games.genre = genre

我遇到上述代码的问题，只有当游戏有评论时才会返回数据，但是当游戏没有我想要的评论时没有返回任何内容

例如，如果游戏没有评论，它仍然应该从该类型返回游戏，但是如果没有评论，则计数为0。当我尝试更改任何东西时，即使有匹配也不会发回任何数据。所以我把它还原回到我上面的工作方法。

Answer 1

尝试将Inner join更改为left join，内部联接仅为您提供匹配记录，而左连接将为您提供左表中的所有行，即使找不到匹配项

SELECT 
  games.id AS id,
  games.ean AS ean,
  games.title AS title,
  games.developer AS developer,
  reviews.NoReviews AS NoReviews 
FROM
  games 
  LEFT JOIN 
    (SELECT 
      ean,
      COUNT(*) AS NoReviews 
    FROM
      reviews 
    GROUP BY ean) reviews 
    ON games.ean = reviews.ean 
WHERE games.genre = genre

Answer 2

将INNER JOIN更改为LEFT OUTER JOIN

查询： -

SELECT games.id AS id,games.ean AS ean, games.title AS title, games.developer AS developer, reviews.NoReviews AS NoReviews
FROM games left outer join
(SELECT ean, COUNT(*) as NoReviews 
FROM reviews GROUP BY ean) reviews 
ON games.ean = reviews.ean 
WHERE games.genre=genre

左外连接将显示左表中的所有记录，以及右表中的匹配记录。

当MYSQL没有匹配时返回计数

2 个答案: