javascript - 谷歌Firestore : Filter documents where sub-key is between given value

Question

我正在开发一款浏览器游戏，我在 firestore 中有一个如下所示的集合:

{
    title: "doc 1",
    requirements: {
        level: {
            min: 2,
            max: 3
        }
    }
},
{
    title: "doc 2",
    requirements: {
        level: {
            min: 6,
            max: 8
        }
    }
},
{
    title: "doc 3",
    requirements: {
        level: {
            min: 8,
            max: 9
        }
    }
}

有什么方法可以查询所有符合给定级别的文档，比如当我的级别为 8 时，我只想获取文档“doc 2”和“doc 3”，因为级别要求匹配？

我试过类似的东西

ref.where("requirements.level.min", "<=", level);
ref.where("requirements.level.max", ">=", level);

我还尝试将文档中的结构更改为:

{
    title: "doc 1",
    requirements: {
        level: [2, 3]
    }
},
{
    title: "doc 2",
    requirements: {
        level: [6, 7, 8]
    }
}

然后像这样过滤

ref.where("requirements.level", "array-contains", level)

但是 firestore 总是返回我所有的文档。

Answer 1

Array contains 在这里似乎是正确的选择。我检查了你的例子 - recreated quite similar structure in my firestore (其他文档具有相同的结构，但级别从 5 到 7)并运行这些查询:

const docs = firebase.firestore().collection('docs');
docs.where('requirements.levels','array-contains',2).get().then((snap) => {
    console.log(snap.docs.length); //<- outputs 1, correct
  });
docs.where('requirements.levels','array-contains',4).get().then((snap) => {
    console.log(snap.docs.length); //<- outputs 0, correct
  });
docs.where('requirements.levels','array-contains',5).get().then((snap) => {
    console.log(snap.docs.length); //<- outputs 1, correct
  });