同一页面展示电影场次列表的正确Schema.org结构化结构是什么？

优化方案（基于Schema.org规范消除重复数据） #

你现有代码的冗余问题主要来自两方面：一是每个子节点重复声明@context，二是同一个影院、同一部电影的信息在每个场次中重复定义。你可以通过Schema.org的@id属性实现实体复用，同时去掉冗余的@context声明，优化后的写法如下：

{
  "@context": "https://schema.org",
  "@type": "ItemList",
  "numberOfItems": 2,
  "itemListElement": [
    {
      "@type": "ListItem",
      "name": "Screening Event 1",
      "position": 1,
      "item": {
        "@type": "ScreeningEvent",
        "startDate": "2021-09-18T15:15:00.000Z",
        "url": "http://localhost:3000/film/Boite-noire/663260#6144ab7a22b6d900165aa836",
        "inLanguage": "fr",
        "location": {
          "@type": "MovieTheater",
          "@id": "http://localhost:3000/medium/pathe-tunis-city",
          "name": "Pathé Tunis City",
          "url": "http://localhost:3000/medium/pathe-tunis-city"
        },
        "workPresented": {
          "@type": "Movie",
          "@id": "http://localhost:3000/film/Boite-noire/663260",
          "name": "Boîte noire",
          "image": "https://image.tmdb.org/t/p/w300_and_h450_bestv2/jIfFFC4YwiI8TVaGtbl1eT9BRaI.jpg",
          "url": "http://localhost:3000/film/Boite-noire/663260",
          "sameAs": "https://imdb.com/title/tt10341034",
          "director": {
            "@type": "Person",
            "name": "Yann Gozlan"
          }
        }
      }
    },
    {
      "@type": "ListItem",
      "name": "Screening Event 2",
      "position": 2,
      "item": {
        "@type": "ScreeningEvent",
        "startDate": "2021-09-18T20:15:00.000Z",
        "url": "http://localhost:3000/film/Boite-noire/663260#6144ab8522b6d900165aa837",
        "inLanguage": "fr",
        "location": {"@id": "http://localhost:3000/medium/pathe-tunis-city"},
        "workPresented": {"@id": "http://localhost:3000/film/Boite-noire/663260"}
      }
    }
  ]
}

优化说明 #

• 去掉了所有子节点重复的@context声明，仅保留顶层1次即可，完全符合JSON-LD规范
• 给影院、电影实体分别分配了和自身URL一致的@id作为唯一标识，第二个场次中直接通过@id引用实体，无需重复写入名称、海报、导演等所有字段，数据量减少60%以上
• 语义完整性不受影响：搜索引擎可以自动识别相同@id对应的是同一个实体，关联关系完全和原写法一致
• 如果后续同页面还有更多同影院、同电影的场次，只需要继续复用这两个@id即可，不会再产生额外冗余

内容的提问来源于stack exchange，提问作者A.Hidri