typescript 有没有办法忽略打字稿中的类型不兼容？
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Is there a way to ignore type incompatibility in typescript?
提问by Christian Nielsen
I have been looking at using typescript with mongoose for MongoDB. Mostly it has been working great, but with certain types of quesites I get warnings from the typescript compiler.
我一直在考虑在 MongoDB 中使用 typescript 和 mongoose。大多数情况下它运行良好，但是对于某些类型的问题，我从打字稿编译器收到警告。
If I do an or like so:
如果我这样做或喜欢这样：
{"$or": [{done: {"$exists": false}}, {done:false}]}
I get the following warning:
我收到以下警告：
Incompatible types in array literal expression: Types of property 'done' of types '{ done: { $exists: bool; }; }' and '{ done: bool; }' are incompatible.
Incompatible types in array literal expression: Types of property 'done' of types '{ done: { $exists: bool; }; }' and '{ done: bool; }' are incompatible.
I understand why, but is there a way to express this so the compiler will accept it?
我明白为什么，但有没有办法表达这一点，以便编译器接受它？
回答by Ryan Cavanaugh
You can type-assert any of the elements to anyto "turn off" type checking:
您可以对任何元素进行类型断言any以“关闭”类型检查：
[<any>{done: {"$exists": false}}, {done:false}]
Or, if you're initializing a variable, you can do something like this:
或者，如果您正在初始化一个变量，您可以执行以下操作：
var n: any[] = [{done: {"$exists": false}}, {done:false}]
