在一条线上推动和取消移动是行不通的。 我究竟做错了什么？

Question

I'm getting an array of ints from an API denoting days.

[0, 7, 30, 356]

I wrote a small function that appends the moment -generated end-date (calculating days from now) but that's not important. The important part is that the second line of my ternary.

On 0 , I don't want to calculate the end-date, but just "no limit" as label and that entry as the last in the array

wanted result:

[{
  code: 7
  label: 'one week'
},
{
  code: 30,
  label: 'one month'
},
{
  code: 356,
  label: 'one year'
},
{
  code: 0,
  label: 'no limit
}]

I wanted to get real fancy and was already kind of proud about the elegant one-liner, but it just doesn't seem to work. It doesn't remove the first entry. entering it in devtools by itself returns the object literal though, so that part seems to be correct.

for (let i = 0, x = lengths.length; i < x; i++) {
  lengths[i]
    ? lengths[i] = //transform the entry here
    : lengths.push(lengths.shift() && {code: lengths[i], label: 'No limit'})
}
return lengths

Can anyone point out what exactly I am did wrong? I want to understand

clarification:

I want to remove 0 from the first array position, transform and add it to the last postition at the same time

edit

thanks to an answer here, i noticed that i used unshift instead of shift to remove the first element.. which of course didn't work

Answer 1

lengths.unshift() returns value , so it adding same value, not the {code: lengths[i], label: 'No limit'}

You can use splice:

lengths.splice(lengths.length-1, 1, {code: lengths[i], label: 'No limit'})

Here remove 1, an element at position lengths.length-1 and add object .

Answer 2

With some ugly code you can do something like this, i am using , operator to shift() and adjust the variables due to changes index because of mutation

 let lengths = [0, 30, 60, 90] for (let i = 0, x = lengths.length; i < x; i++) { lengths[i]? lengths[i] = {code: lengths[i], label: "dummy"}: (lengths.push({code: lengths[i], label: 'No limit'}), lengths.shift(), i--, x--) } console.log(lengths)

---

Close to your original code,

1. ! need before shift as the return value of shift is 0 ( in our case ) so to make && operator return object we need to make it truthy value
2. Need to adjust indexes else we miss out elements

 let lengths = [0, 30, 60, 90] for (let i = 0, x = lengths.length; i < x; i++) { lengths[i]? lengths[i] = {code: lengths[i], label: "dummy"}: (lengths.push(.lengths:shift() && {code, 0: label, 'No limit'}), i--.x--) } console.log(lengths)

PS:- 1st point can be avoided by using || operator as nicely pointed out in nina's answer

Answer 3

You need a logical OR || instead of a logical AND && , because shift returns zero and this is falsy. The following part in never evaluated this zero is pushed.

lengths.push(lengths.shift() || { code: lengths[i], label: 'No limit' })
//                           ^^

---

You could iterate from the end and shift at the end where item is left.

 var lengths = [0, 7, 30, 356], i = lengths.length; while (i--) { if (lengths[i]) lengths[i] = { code: lengths[i], label: '...' }; else { lengths.shift(); lengths.push({ code: 0, label: 'no limit' }); } } console.log(lengths);

A short approach is to shift and push before you change the values.

 var lengths = [0, 7, 30, 356]; lengths.push(lengths.shift()); console.log(lengths);