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How to store timestamp a JSON object?


How to store timestamp a JSON object?

By : Rhys Price
Date : November 25 2020, 09:00 AM
wish of those help Here's a very simple example: http://plnkr.co/edit/C4sgMK8mg4wZUhiXW58n
We use ngClick in the DOM to invoke a function that lives in our controller, and once this function fires, it creates a new Date object, and adds it to our dates array. The DOM is set up to display all items in the dates array:
code :
app.controller('MainCtrl', function($scope) {
  $scope.dates = [];
  $scope.createNewDate = function() {
    $scope.dates.push(new Date());
  }
});
  <body ng-controller="MainCtrl">
    <button ng-click="createNewDate()">New Date</button>
    <button ng-click="dates=[]">Clear</button>
    <ul>
      <li ng-repeat="date in dates">{{date | date:'M/d/yy h:mm:ss:sss a'}}</li>
    </ul>
  </body>


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Store date as unix timestamp or TIMESTAMP data type in MySQL?

Store date as unix timestamp or TIMESTAMP data type in MySQL?


By : user3542137
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Usually it does not matter whether you use TIMESTAMP or DATETIME datatype.
In older versions, TIMESTAMP was 4 bytes and DATETIME was 8. Think of DATETIME as a picture of a clock; think of TIMESTAMP as an instant in time, worldwide. That is, if you connect to the same database, but from a different timezone, a DATETIME will look the the same, but a TIMESTAMP will be adjusted for timezone. NOW(), SELECTing into PHP, etc, are compatible with both. Both are externally seen as a string, such as '2015-04-25 17:09:01'. Since TIMESTAMP is stored as a 32-bit integer (but you don't see that), it is limited to ~1970-2038. Since DATETIME is clock time, there will be a missing/extra hour twice a year if you switch to/from daylight savings time.
How to let Timestamp object show in type long in JSON?

How to let Timestamp object show in type long in JSON?


By : Tommi Jacky Jones
Date : March 29 2020, 07:55 AM
it should still fix some issue Use .getTime() to convert in to long.
If you are getting an object in variable "created" then convert it into long :
code :
created.getTime();
What format should I use to store timestamp in JSON for mongodb queries used in nodejs?

What format should I use to store timestamp in JSON for mongodb queries used in nodejs?


By : François Chevallier
Date : March 29 2020, 07:55 AM
it helps some times The following answer suggests that the best way is to use the JavaScript Date (constructed also via ISODate), which is native to Mongo. Digging into the BSON documentation it further clarifies the internal values:
JS - JSON: adding a keypair in each object by calculating the difference in time now and a timestamp in the JSON object

JS - JSON: adding a keypair in each object by calculating the difference in time now and a timestamp in the JSON object


By : Matt Ferancini
Date : March 29 2020, 07:55 AM
this will help I know you're aware of the forEach method, but with that, you're making changes to the original array, if you'd like make a new copy of the array, you can use map. Then of course to make 'clones' of the objects within the array by making use of destructuring.
Edit
code :
const array = [{"id": 1,"timestamp": 1545311087282}, {"id": 2, "timestamp": 1545311148254}];
const now = new Date().getTime();

// New copy.
const newArray = array.map(o => ({...o, diff: now - o.timestamp}));

console.log(array);
console.log(newArray);

// Changes original. 
array.forEach(o => o.diff = now - o.timestamp);
console.log(array);
const array = [{"id": 1,"timestamp": 1545311087282}, {"id": 2, "timestamp": 1545311148254}];
const now = new Date().getTime();

// Difference in minutes.
const diff = time => new Date(now).getMinutes() - new Date(time).getMinutes();

// New copy.
const newArray = array.map(o => ({...o, minsDiff: diff(o.timestamp)}));

console.log(array);
console.log(newArray);
Get json object that has latest timestamp using jq

Get json object that has latest timestamp using jq


By : user2914805
Date : March 29 2020, 07:55 AM
I hope this helps you . Simply sort by .createdDate and (assuming you only want one value even if there is more than one with the greatest .createdDate value), select the last one:
code :
.items
| sort_by(.createdDate)[-1].description
.items
| sort_by(.createdDate)
| (.[-1].createdDate) as $max
| .[]
| select($max == .createdDate)
| .description
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