php - 如何将原始 SQL 查询转换为 Silverstripe SQLQuery 抽象层

Question

我有一个页面，我正在尝试从数据库中提取与该页面相关的文章。我有提取我需要的内容的 SQL 查询，但我不断收到错误消息“'where 子句'中的未知列'Fashion'”。我相信我需要将它从，

$FilteredStories =  DB::query(' SELECT C.ID, C.URLSegment, C.Title, B.Title AS "Category"
                                FROM `articlepage_categories` AS A
                                JOIN articlecategory AS B ON A.ArticleCategoryID = B.ID
                                JOIN sitetree AS C ON A.ArticlePageID = C.ID
                                WHERE B.Title = "Fashion" LIMIT 5')
                    ->value();

进入 SQLQuery 抽象层，但我不知道如何。有人可以告诉我如何创建具有多个连接的 SQLQuery 抽象层吗？

注意事项

• 我使用的是 Silverstripe 3.6.1 版
• “时尚”目前是硬编码的，但将被替换为我将传入的变量。

Answer 1

SilverStripe的数据库默认使用ANSI sql_mode，其中字符串字面量需要用单引号括起来。您需要将 "Fashion" 周围的双引号替换为单引号，如下所示:

$FilteredStories =  DB::query('SELECT C.ID, C.URLSegment, C.Title, B.Title AS "Category"
                            FROM `articlepage_categories` AS A
                            JOIN articlecategory AS B ON A.ArticleCategoryID = B.ID
                            JOIN sitetree AS C ON A.ArticlePageID = C.ID
                            WHERE B.Title = \'Fashion\' LIMIT 5')

此处转义，因为外引号是单引号。

您的查询将用 SQLSelect 表示，如下所示:

$filteredStories = SQLSelect::create();
$filteredStories->selectField('"sitetree"."ID", "sitetree"."URLSegment", "sitetree"."Title", "articlecategory"."Title" AS "Category"');
$filteredStories->setFrom('articlepage_categories');
$filteredStories->addLeftJoin('articlecategory', '"articlecategory"."ID" = "articlepage_categories"."ArticleCategoryID"');
$filteredStories->addLeftJoin('sitetree','"sitetree"."ID" = "articlepage_categories"."ArticlePageID"');
$filteredStories->addWhere('"articlecategory"."Title" = \'Fashion\'');
$filteredStories->setLimit(5);