'JSON array of objects with properties “from” and “to” and group it by time period
In case if dates in the array are intersect then function should combine them in one period, else periods should be separated
var dates = [
{
"from":"03/01/2021",
"to":"03/05/2021"
},
{
"from":"03/08/2021",
"to":"03/10/2021"
},
{
"from":"03/07/2021",
"to":"03/20/2021"
}
];
var item = dates;
var index = 0;
var out = document.querySelector('#test');
for(var i = 0; i < item.length; i++){
index++;
var from_i = Date.parse(item[i].from);
var to_i = Date.parse(item[i].to);
var from_index = Date.parse(item[index].from);
var to_index = Date.parse(item[index].to);
if(from_index >= from_i && from_index <= to_i){
let updated_from = new Date(from_index);
let update_to = new Date(to_index);
console.log(updated_from);
out.innerHTML += updated_from.toDateString() + update_to.toDateString() + "<br/>";
}
}
<div id="test"> </div>
Case of incoming jSON: [ { "from":"03/01/2021", "to":"03/06/2021" }, { "from":"03/10/2021", "to":"03/15/2021" }, { "from":"03/20/2021", "to":"03/25/2021" } ]
Output: Mar 1-5, 7-20
Can some help? Learning JS…little bit hard for me(
Solution 1:[1]
you can do something like this
const data = [ { "from":"03/01/2021", "to":"03/06/2021" }, { "from":"03/10/2021", "to":"03/15/2021" }, { "from":"03/20/2021", "to":"03/25/2021" } ]
const monthToString = month => {
const months = {
'01': 'Jan',
'02': 'Feb',
'03': 'Mar',
'04': 'Apr',
'05': 'May',
'06': 'Jun',
'07': 'Jul',
'08': 'Aug',
'09': 'Sep',
'10': 'Oct',
'11': 'Nov',
'12': 'Dec',
}
return months[month] || month
}
const result = Object.entries(data.reduce((res, {from, to}) => {
const [monthFrom, day, year] = from.split('/')
const [m, dayTo, y] = to.split('/')
const month = monthToString(m)
return {
...res,
[month]: [...(res[month] || []), [day, dayTo]]
}
}, {})).map(([month, days]) => `${month} ${days.map(([from, to]) => `${parseInt(from)}-${parseInt(to)}`).join(', ')}`).join('\n')
console.log(result)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | R4ncid |
