從舊表創建部落格新表

我正在從舊表重新創建數據庫結構。新表應該優化並且易於理解。

在舊數據庫中，開發人員使用video表名來儲存影片部落格。但是我不認為它是儲存影片部落格的正確名稱。下面是舊表

video:
   _attributes: { phpName: Video }
   ID: { type: INTEGER, size: '11', primaryKey: true, autoIncrement: true, required: true }
   coupon_banner_id: { type: INTEGER, size: '10', foreignTable: coupon_banner, foreignReference: id, onUpdate: CASCADE, onDelete: CASCADE }
   blog: { type: TINYINT, size: '1', required: true, defaultValue: '0' }
   catid: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   title: { type: VARCHAR, size: '100', required: true, defaultValue: '' }    
   description: { type: LONGVARCHAR, required: true }
   blogtopid: { type: INTEGER, size: '11', required: true, defaultValue: '1' }
   video: { type: VARCHAR, size: '50', required: true, defaultValue: '' }
   vlink: { type: LONGVARCHAR, required: true }
   picture: { type: VARCHAR, size: '255', required: true, defaultValue: '' }
   video_picture_alt: { type: VARCHAR, size: '255', required: false, defaultValue: '' }
   apicture: { type: VARCHAR, size: '150', required: true, defaultValue: '' }
   usid: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   datein: { type: TIMESTAMP, required: true, defaultValue: '0000-00-00 00:00:00' }
   koview: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   korating: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   kototrat: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   konota: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   kocomm: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   kofavorite: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   kofeatured: { type: INTEGER, size: '11', required: true, defaultValue: '0' }
   koorder: { type: INTEGER, size: '11', required: true, defaultValue: '999' }
   commtime: { type: TIMESTAMP, required: true, defaultValue: '0000-00-00 00:00:00' }
   is_active: { type: BOOLEAN, required: true, defaultValue: '0'}
   slug: { type: VARCHAR, size: '100', required: true, defaultValue: '' }
 video_cat:
   _attributes: { phpName: VideoCat }
   ID: { type: INTEGER, size: '11', primaryKey: true, autoIncrement: true, required: true }
   category: { type: VARCHAR, size: '50', required: true, defaultValue: '' }
   meta_title: { type: VARCHAR, size: '255', required: true, defaultValue: '' }
   category_desc: { type: LONGVARCHAR, required: false, defaultValue: '' }
   meta_desc: { type: LONGVARCHAR, required: false, defaultValue: '' }
   position: { type: INTEGER, size: '3', required: true, defaultValue: '999' }

我將其重命名並優化如下：

article:
   id,
   category_id,
   user_id,
   title,
   slug,
   description,
   video_link,
   video_description,
   video_thumbnail,
   Video_thumbnail_alt,
   article_description ( I will be using CKEditor here so styles, which data type is better text or longvarchar? ),
   is_active (int or boolean which one is better),
   is_featured (int or boolean which one is better),
   is_favorite (int or boolean which one is better),
   created_at DATETIME,
   updated_at DATETIME

您認為我的新優化表還有哪些需要改進的地方？我還想知道將影片和文章儲存在單獨的表中是否更好，或者可以像我一樣儲存在同一個表中？

是否單獨的表 - 這取決於

• 大多數列是否相同？
• 表上的大多數操作是否相同？

如果一個影片可以屬於多個類別，您將需要一個多對多映射表。

什麼是position，如何維護？

引用自：https://dba.stackexchange.com/questions/224312