Firestore 选择不为空的地方 [英] Firestore select where is not null

问题描述

我正在使用 firebase 来管理我的项目，但我无法使用 where 子句创建查询，其中某些值不为空.

I'm using firebase to manage my project and I cannot get to create a query with a where clause where some value is not null.

示例:我有一组员工.每个都有一个设备列表作为对象，其中键是设备 ID，值是颜色.

Example: I have a collection of employees. Each have a list of equipments as an object where the key is the equipment id and the value a color.

user = {
    firstName: 'blabla',
    lastName: 'bloblo',
    equipments: {
        123: 'blue',
        124: 'red'
    }
}

我想在设备中得到所有拥有某种设备的员工.让我们说 123.

I would like to get all the employees who has a certain equipment in the equipments. Lets say 123.

这里是 Select * from Employees ，其中 devices.123 不为空.我试过了:

It comes to Select * from Employees where equipments.123 is not null. I've tried:

firestore.collection('employees').where(`equipments.${equipmentId}`, '!=', null)

但它不起作用.

我似乎无法让它工作.你能帮我吗.

I can't seems to make it work. Can you help me.

推荐答案

2020 年 9 月更新: v7.21.0 引入了对不等于 (!=) 查询的支持！这意味着您现在可以使用以下代码:

Update Sep 2020: v7.21.0 introduces support for not equals (!=) queries! That means you can now use the code bellow:

firestore.collection('employees').where(`equipments.${equipm‌​entId}`, '!=', null)

---

上一个答案:

Firestore 没有不等于"操作员.但是从逻辑上看，您要做的是查询String 而不是null 的值.Firestore 可以做到这一点，前提是您将字符串传递给 where() 函数.

Firestore has no "not equal" operator. But looking at the logic, what you're trying to do is query for values which are String, and not null. Firestore can do that if you pass a String to the where() function.

所以你可以做的是查询低于 uf8ff 的值.这是 Unicode 范围内的一个非常高的代码点.由于它位于 Unicode 中大多数常规字符之后，因此此查询将返回 String 类型的所有内容:

So what you can do is query for values lower than uf8ff. That's a very high code point in the Unicode range. Since it is after most regular characters in Unicode, this query will return everything that is of type String:

firestore.collection('employees').where(`equipments.${equipm‌​entId}`, '<', 'uf8ff')

或者您可以简单地查询高于"的值.(空String):

Or you can simply query for values higher than "" (empty String):

firestore.collection('employees').where(`equipments.${equipm‌​entId}`, '>', '')

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