CODEWITHSUNDEEP
javascripthtmlangularjsdatetime
hassanzi
hassanzi

Reputation: 211

collect numbers and convert them to date in angularjs

I have 3 scopes in angularjs, one for year, one for month and one for day coming from 3 input boxes. I want to combine them as date, the following code insert the date correctly but with 22:00:00 added to it like 2019-06-01 22:00:00.000. How to insert the date only without time

HTML

<div class="col-md-1">
    <div class="form-group has-error">
        <label class="title_lable">Year:</label>
        <input id="code" type="number" class="form-control input-sm"
               ng-model="year">
    </div>
</div>
<div class="col-md-2">
    <div class="form-group has-error">
        <label class="title_lable">Latin Description:</label>
        <input type="text" class="form-control input-sm"
               ng-model="latindesc">
    </div>
</div>
<div class="col-md-2">
    <div class="form-group">
        <label class="title_lable">Local Description:</label>
        <input type="text" class="form-control input-sm"
               ng-model="localdesc">
    </div>
</div>
<div class="col-md-1">
    <div class="form-group">
        <label class="title_lable">Month:</label>
        <select ng-show="lang==1" class="form-control input-sm"
                ng-model="month"
                ng-options="m.mnthid as m.mnthanm for m in months">
        </select>
        <select ng-show="lang==0" class="form-control input-sm"
                ng-model="month"
                ng-options="m.mnthid as m.mnthenm for m in months">
        </select>
    </div>
</div>
<div class="col-md-1">
    <div class="form-group">
        <label class="title_lable">Day:</label>
        <input type="number" class="form-control input-sm" ng-model="day">
    </div>
</div>

Angular

var dFrom = new Date($scope.month + '/' + ($scope.day + 1) + '/' + $scope.year );
        var days = $scope.ndays;
        console.log(days);
        var todate = new Date($scope.month + '/' + ($scope.day +1) + '/' + $scope.year);
        todate = addDays(todate, days - 1);
        console.log(todate);

        var vacation = {
            Sys_Key: $scope.syskey,
            Vac_Code: $scope.code,
            L_Desc: $scope.latindesc,
            A_Desc: $scope.localdesc,
            V_Start_Date: $scope.day,
            V_Month: $scope.month,
            V_Days_Count: $scope.ndays,
            V_Year: $scope.year,
            From_Date: dFrom,
            To_Date: todate
        }

function addDays(startDate, numberOfDays) {
    var returnDate = new Date(
        startDate.getFullYear(),
        startDate.getMonth(),
        startDate.getDate() + numberOfDays,
        startDate.getHours(),
        startDate.getMinutes(),
        startDate.getSeconds()
    );
    return returnDate;
}

Upvotes: 2

Views: 878

Answers (3)

georgeawg
georgeawg

Reputation: 48948

the code insert the date correctly but with 22:00:00 added to it like 2019-06-01 22:00:00.000. How to insert the date only without time

Under-the-hood the date is parsed to 00:00 hours UTC time. The standard .toString method displays it as local time.

var d = new Date("2019-06-01");

console.log(d.toString());

console.log(d.toISOString());

If you want it parsed to local time append "T00:00"

var d = new Date("2019-06-01T00:00");

console.log(d.toString());

console.log(d.toISOString());

When the time zone offset is absent, date-only forms are interpreted as a UTC time and date-time forms are interpreted as local time.

For more information, see

Upvotes: 1

Manish Khedekar
Manish Khedekar

Reputation: 402

You can do it as,

let dateString = "05/05/1997";

let date = new Date(dateString);

console.log(date.toLocaleString('en-US', {
      day:"2-digit",
      month:"2-digit",
      year:"numeric"
    }));

Please checkout following link,

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date/toLocaleString

Upvotes: 0

Scott Flodin
Scott Flodin

Reputation: 330

When adding in javascript, be sure you are using numerical values and not strings. You can convert strings to integers with parseInt(string)

Upvotes: 1

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