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

我有一个页面，我试图从数据库中提取与该页面相关的文章。我有一个SQL查询，它提取了我需要的内容，但我不断得到错误“where子句”中的“Unknown column'Fashion”。我想我需要把它从

$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。有人能告诉我如何创建具有多个连接的SQLQuery抽象层吗

注释

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

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SilverStripe的数据库默认使用

ANSI

sql\u模式，其中字符串文本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，因为外部引号是单引号

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);