Question

我有一个数组，其中包含工作日的对象，我希望按＆＃34;打开＆＃34;或者＆＃34;关闭＆＃34; （不要让我们在最后一个阵列中存在）。

let array = [
             [
               {"weekday":1,"opens":"09:00","closes":"11:00"}, 
               {"weekday":1,"opens":null,"closes":null}
             ],
             [
               {"weekday":2,"opens":"09:00","closes":"11:00"}, 
               {"weekday":2,"opens":"12:30","closes":"17:00"},
               {"weekday":2,"opens":"18:00","closes":"null"}
             ], ...
           ]

我想返回一个新创建的数组，以便我不会改变原始数组。

我目前的解决方案看起来很像但感觉很难看

let newArray = [];

array.forEach( (day, index)  => {
    day = day.filter( timeblock => 
       timeblock.opens != null && timeblock.closes != null
    );
    newArray.push(day);
});

如何过滤嵌套数组更优雅？（如果需要，jsfiddle：https://jsfiddle.net/2jukvsoy/1/）

Answer 1

let newArray = array.map(day => 
    day.filter(timeblock => 
        timeblock.opens != null && timeblock.closes != null
    )
);

Answer 2

没什么可做的，但也许你可以使用map而不是forEach

let array = [
             [
               {"weekday":1,"opens":"09:00","closes":"11:00"}, 
               {"weekday":1,"opens":null,"closes":null}
             ],
             [
               {"weekday":2,"opens":"09:00","closes":"11:00"}, 
               {"weekday":2,"opens":"12:30","closes":"17:00"},
               {"weekday":2,"opens":"18:00","closes":"null"}
             ]
           ]

const newArray = array.map(day => {
    return day.filter(timeblock => timeblock.opens && timeblock.closes);
});

console.log(newArray)

通过删除不需要的嵌套对象属性来过滤对象数组

2 个答案: