我已经定义了一个函数，它递归地遍历嵌套对象（或非嵌套对象）并查找特定键，提取其值

const findName = <T extends object>(obj: T, keyToFind: string): T[] => {
   return Object.entries(obj)
      .reduce((acc, [key, value]) => {
         if(key === keyToFind) {
            return acc.concat(value)
         } else {
            //is the next level of nesting an object so we can keep recursively searching for the key?
            if (typeof value === 'object') {
               return acc.concat(findName(value, keyToFind)) //problem is here because value is of type any
            } else {
               return acc
            }
         }
      }, [])
}

我正在尝试为value定义一个类型，因为当我在递归步骤中调用findName时，我得到了以下错误

No overload matches this call.
  Overload 1 of 2, '(...items: ConcatArray<never>[]): never[]', gave the following error.
    Argument of type 'any[]' is not assignable to parameter of type 'ConcatArray<never>'.
      The types returned by 'slice(...)' are incompatible between these types.
        Type 'any[]' is not assignable to type 'never[]'.
          Type 'any' is not assignable to type 'never'.
  Overload 2 of 2, '(...items: ConcatArray<never>[]): never[]', gave the following error.
    Argument of type 'any[]' is not assignable to parameter of type 'ConcatArray<never>'.

我不确定如何定义一个类型，所以编译器不再抱怨这个问题。
这是当前场景的Playground
谢谢

• • 更新**

我通过将reduce的泛型类型设置为T[]，成功地对该错误进行了排序，但value仍然是any类型，因此开始出现linting，是否有办法在连接时将其定义为字符串

const findName = <T extends object>(obj: T, keyToFind: string): T[] => {
   return Object.entries(obj)
      .reduce<T[]>((acc, [key, value]) => {
         if(key === keyToFind) {
            return acc.concat(value)
         } else {
            //is the next level of nesting an object so we can keep recursively searching for the key?
            if (typeof value === 'object') {
               return acc.concat(findName(value, keyToFind))
            } else {
               return acc
            }
         }
      }, [])
}