我认为 Splice(remove) 方法的工作原理与其实际工作完全相反

Question

我希望只删除 "deleteNumber" 方法中的给定索引。但它做的恰恰相反。它删除除给定索引之外的所有索引。

Javascript代码：

deleteNumber = (index) => {
    console.log("index: ", index);
    this.setState((state) => {
        const numbers = state.numbers.splice(index, 1);
        this.storeData('numbers', JSON.stringify(numbers));
        return(
        {
            numbers: numbers,
        }
        )
    })
}

Answer 1

拼接改变给定的数组和returns它删除的元素。像这样的东西会将状态设置为新的（改变的）数组。

deleteNumber = (index) => {
  let numbers = this.state.numbers;
  numbers.splice(index, 1);
  this.setState({
    numbers
  });
}

我认为 Splice(remove) 方法的工作原理与其实际工作完全相反

I think Splice(remove) method works as exactly opposite of its actual work

javascript

react-native

splice