Typescript错误，因为类型没有重叠，所以此条件将始终返回"true" [英] Typescript error This condition will always return 'true' since the types have no overlap

问题描述

我在表单组中有此条件:

I having this condition on a form group:

if((age>17 && (this.frType=="Infant")) 
|| (age>40 && this.frType=="Grandchild")
|| (age<=5 && 
   (this.frType!="Child" 
   || this.frType!="Infant" 
   || this.frType!="Grandchild" || this.frType!="Cousin")))

它包含3个主要条件:

1. 如果17岁的人无法设置为infant
2. 如果一个人大于40岁，那么他就不能成为grandchild
3. 如果一个人的年龄小于5岁，则应为child，infant，grandchild或cousin.

1. If a person aged 17, cannot be set to infant
2. If a person is bigger than 40, he cannot be a grandchild
3. If a person is less than 5 years, he should be child, infant, grandchild or cousin.

如果满足以下条件之一，我将发送一条错误消息.

If one of these conditions is true, I will send an error message.

我收到的错误是:

[ts]由于类型 '"Child"'和'"Infant"'没有重叠. [2367]

[ts] This condition will always return 'true' since the types '"Child"' and '"Infant"' have no overlap. [2367]

关于if条件的这一部分`:

On this part of the if condition`:

|| this.frType!="Infant" || this.frType!="Grandchild" || this.frType!="Cousin")))

我在其他组件中使用的是确切条件，并且没有显示错误.

I am using the exact condition in a different component, and it does not show an error.

if((age>17 && (this.family_relation_type=="Infant")) 
|| (age>40 && this.family_relation_type=="Grandchild")
|| (age<=5 && 
   (this.family_relation_type!="Child" || 
    this.family_relation_type!="Infant" || 
    this.family_relation_type!="Grandchild" || 
    this.family_relation_type!="Cousin")))

这是我计算这两个部分的年龄的方法:

Here is how I am calculating the age in both components:

let timeDiff = Math.abs(Date.now() - this.formGroup.controls['dob'].value);
let age = Math.floor((timeDiff / (1000 * 3600 * 24))/365);

推荐答案

考虑独立表达式:

(this.frType!="Child" || this.frType!="Infant")

如果frType为Child，则第二部分为true，因此表达式的计算结果为true.如果frType是Infant，则第一部分为true，因此表达式的计算结果为true.如果frType既不是 Child也不是Infant，则第一部分将为true，并且表达式将再次求值到true-逻辑错误，它将始终解析为<​​c9>.

If frType is Child, the second part will be true, so the expression will evaluate to true. If frType is Infant, then the first part will be true, so the expression will evaluate to true. If frType is neither Child nor Infant, then the first part will be true, and the expression will, again, evalute to true - the logic is faulty, it'll always resolve to true.

(如果为Grandchild和Cousin添加其他||条件，同一件事不断发生-它将始终解析为<​​c9>)

(If you add additional || conditions for Grandchild and Cousin, the same thing keeps happening - it'll always resolve to true)

请改用&&:

|| (age<=5 && (
   this.frType!="Child" 
   && this.frType!="Infant" 
   && this.frType!="Grandchild"
   && this.frType!="Cousin"
 ))

或者，为了使逻辑更容易理解，您可以考虑使用数组，然后使用.includes:

Or, to make the logic easier to follow, you might consider using an array, and use .includes:

const kidsFiveAndUnder = ['Child', 'Infant', 'Grandchild', 'Cousin'];
// ...
|| (age <= 5 && !kidsFiveAndUnder.includes(this.frType))

这篇关于Typescript错误，因为类型没有重叠，所以此条件将始终返回"true"的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！