Question

我有以下查询：

SELECT e_c.*, c.name, j.status, j.version, j.articleId, j.title FROM assetcategory AS c
    INNER JOIN assetentries_assetcategories AS e_c 
        ON c.categoryId = e_c.categoryId AND c.name = 'news'
    INNER JOIN assetentry AS e
        ON e.entryId = e_c.entryId
    INNER JOIN journalarticle AS j
        ON j.resourcePrimKey = e.classPK
        AND e.classNameId = (SELECT classNameId FROM classname_ WHERE value = 'com.liferay.portlet.journal.model.JournalArticle')
        AND j.companyId= e.companyId
WHERE j.status = 0

返回news中的所有类别journalarticles。从结果中，我需要为每个articleId选择最新版本。例如，假设有一篇文章有4个版本，即使标题不同，它也是同一篇文章，因为它将具有相同的articleId。因此，对于每个唯一的articleId，我需要最新的version。我该怎么办？

Answer 1

将联接添加到子查询中，该子查询可查找每篇文章的最新版本：

SELECT e_c.*, c.name, j1.status, j1.version, j1.articleId, j1.title
FROM assetcategory AS c
INNER JOIN assetentries_assetcategories AS e_c 
    ON c.categoryId = e_c.categoryId AND c.name = 'news'
INNER JOIN assetentry AS e
    ON e.entryId = e_c.entryId
INNER JOIN journalarticle AS j1
    ON j1.resourcePrimKey = e.classPK AND
       e.classNameId = (SELECT classNameId FROM classname_
                   WHERE value = 'com.liferay.portlet.journal.model.JournalArticle') AND
       j.companyId = e.companyId
INNER JOIN
(
    SELECT articleId, MAX(version) AS max_version
    FROM journalarticle
    WHERE status = 0
    GROUP BY articleId
) j2
    ON j1.articleId = j2.articleId AND j1.version = j2.max_version;

上面加入别名为j2的子查询的联接的基本思想是将结果集限制为每篇文章的最新版本。我们不必更改其余查询。

从先前查询的结果中提取文章的最新版本

1 个答案: