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按键对对象进行排序,同时保留当前值以匹配数组顺序

[英]Sort object by key whilst retaining present values to match order of array

I have two objects.我有两个对象。 One in is in the format (footerMenuOptions) :格式之一是(footerMenuOptions)

[{Home: true}, {About: false}, {Features: false}, {Contact: false}] 

The second is in the format (this.navbarMenuOptions) :第二个是格式(this.navbarMenuOptions)

["Home", "About", "Features", "Contact"]

At times, the order of the second object (this.navbarMenuOptions) will change, to lets say:有时,第二个对象(this.navbarMenuOptions)的顺序会改变,比如说:

["About", "Home", "Features", "Contact"]

I want the first objects order (footerMenuOptions) to change to be able to reflect this (ie the keys), but the values to remain intact).我希望第一个对象顺序(footerMenuOptions)改变以能够反映这一点(即键),但值保持不变)。

In the case where the key-value pair does not exist, it should just be created with a default value of false (ie if this.navbarMenuOptions has a new entry).在键值对不存在的情况下,应该使用默认值 false 创建它(即如果this.navbarMenuOptions有一个新条目)。

What would be the most performant way of accomplishing this:实现这一目标的最有效方法是什么:

Code is as follows:代码如下:

toggleFooterMenuOptionVisibility(index: number) {
   // Some other stuff happens beforehand which isn't important
   footerMenuOptions = this.sortFooterMenuOptions(footerMenuOptions);
}

sortFooterMenuOptions(footerMenuOptions) {
   // this.navbarMenuOptions is a global variable so it doesn't need to be passed in
   return footerMenuOptions;
}

You can use your main object's keys and get indexes from second array and sort your object's keys by these indexes like below:您可以使用主对象的键并从第二个数组中获取索引,并通过这些索引对对象的键进行排序,如下所示:

 var arr1 = { Home: true, About: false, Features: false, Contact: false } var arr2 = ["About", "Home", "Features", "Contact"]; const ordered = {}; Object.keys(arr1).sort((a, b) => { return arr2.indexOf(a) - arr2.indexOf(b); }).forEach(r => ordered[r] = arr1[r]); console.log(ordered);

I assumed your object has 1 key or the first key is the same with values in the second array.我假设您的对象有 1 个键,或者第一个键与第二个数组中的值相同。 So if we apply it to your example it should become like this :所以如果我们把它应用到你的例子中,它应该变成这样:

 sortFooterMenuOptions(unsortedFooterMenuOptions) { const sortedFooterMenuOptions = {}; Object.keys(unsortedFooterMenuOptions) .sort((a, b) => { return this.navbarMenuOptions.indexOf(a) - this.navbarMenuOptions.indexOf(b); }).forEach(r => sortedFooterMenuOptions[r] = unsortedFooterMenuOptions[r]); return sortedFooterMenuOptions; }

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