如何在 TypeScript 中使用 Spread 运算符组合对象

Question

我需要使用 Spread 运算符向对象添加值。但是当我使用它时，以前的值将从对象中删除。

我的对象是这样的

let newObj = {
name: 'sachin',
lastname: 'Tendulkar',
full_name: 'Sachin Ramesh Tendulkar',
test: {
  test_debut: '15 November 1989 v Pakistan'
},
odi: {
  odi_debut: '18 December 1989 v Pakistan'
},
t20: {
  t20_debut: '1 December 2006 v South Africa'
}
}

我需要为此对象添加更多值

{
    name: 'sachin',
    lastname: 'Tendulkar',
    full_name: 'Sachin Ramesh Tendulkar',
    Born: '24 April 1973',
    test: {
        debut: '15 November 1989 v Pakistan',
        last_match: '14 November 2013 v West Indies',
        cap: 187
    },
    odi: {
        debut: '18 December 1989 v Pakistan',
        last_match: '18 March 2012 v Pakistan',
        cap: 74
    },
    t20: {
        debut: '1 December 2006 v South Africa',
        last_match: '1 December 2006 v South Africa',
        cap: 11
    }
}

所以我添加了

const newValues = {
  Born: '24 April 1973',
  test: {
            last_match: '14 November 2013 v West Indies',
            cap: 187
        },
   odi: {
            last_match: '18 March 2012 v Pakistan',
            cap: 74
        },
   t20: {
            last_match: '1 December 2006 v South Africa',
            cap: 11
        }
}

当我使用

newObj = {...newObj , ...newValues };

我得到的回复低于

{
    name: 'sachin',
    lastname: 'Tendulkar',
    full_name: 'Sachin Ramesh Tendulkar',
    Born: '24 April 1973',
    test: {
        last_match: '14 November 2013 v West Indies',
        cap: 187
    },
    odi: {
        last_match: '18 March 2012 v Pakistan',
        cap: 74
    },
    t20: {
        last_match: '1 December 2006 v South Africa',
        cap: 11
    }
}

如何在 typeScript 中解决这个问题

Answer 1

您需要单独 select 对象属性并单独展开其他对象属性

const obj[key] = [...newObj[key], ...newValues[key]]

截至目前，它只是用新属性覆盖您的属性，因为它不会比较它覆盖的键

如何在 TypeScript 中使用 Spread 运算符组合对象

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