javascript - 使用尾随对象的键和递减键删除对象对象中的键

Question

这很难用语言解释，所以让我向您展示我所拥有的并从那里开始工作。

我有以下对象:

{
  1: { name: 'Initial Step 1', stepNumber: 1 },
  2: { name: 'Initial Step 2', stepNumber: 2 },
  3: { name: 'Initial Step 3', stepNumber: 3 },
  4: { name: 'Initial Step 4', stepNumber: 4 },
}

我想为此步骤执行删除功能，然后减少尾随键，以便获得以下结果:

{
  1: { name: 'Initial Step 1', stepNumber: 1 },
  2: { name: 'Initial Step 3', stepNumber: 2 },
  3: { name: 'Initial Step 4', stepNumber: 3 },
};

我只是不确定实现我的 deleteStep 函数的最佳方式。

我已经实现如下:

export const deleteStep = (stepNumber, steps) => {
    // What should I do here?
    return steps;
};

谢谢!

Answer 1

您可以删除实际键并检查是否存在更多键，然后创建一个新属性并删除旧引用。

function deleteAndShift(object, key) {
    delete object[key];
    while (++key in object) {
        object[key].stepNumber--;
        object[key - 1] = object[key];
        delete object[key];
    }
}

var object = { 1: { name: 'Initial Step 1', stepNumber: 1 }, 2: { name: 'Initial Step 2', stepNumber: 2 }, 3: { name: 'Initial Step 3', stepNumber: 3 }, 4: { name: 'Initial Step 4', stepNumber: 4 } };

deleteAndShift(object, 2);

console.log(object);

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