CODEWITHSUNDEEP
javascriptjquerytimeago
comebal
comebal

Reputation: 946

How to make timeago.js show only hours ago

I am using timeago.js to show the date. However, I want to show the "hours ago" feature of the plugin if a particular date is less than 2 days. If a date is more than 2 days, it'll show the usual "3 days ago".

Thanks for the help in advance.

Upvotes: 1

Views: 787

Answers (3)

atool
atool

Reputation: 496

I think timeago.js can help you, and the tiny library is supported by me.

You can just customize you locale function for your need.

const yourLocale = (number, index, totalSec) => {
   // format totalSec into hours.
   const hours = getHours(totalSec);
   return hours +' hours ago';
};

Then register it.

timeago.register('yourLocale', yourLocale);

Them format use it.

timeago.format(time, 'yourLocale');

done!

Upvotes: 0

jave.web
jave.web

Reputation: 15032

You can of course modify plugin like you did, or you could run the dates over again and check whether they are more than two days or not - if they are - just rewrite the string to hours, something like this:

$(function(){
    var oneHour = 1*60*60*1000; // one hour in miliseconds
    var twoDays = 2*24*oneHour; // two days in miliseconds

    $('.timeAgoSelector').each(function(){
        $t = $(this);
        var agoDate = new Date( $t.attr('title') ).getTime()
        var currDate = new Date().getTime();
        var timeAgo = currDate - agoDate;
        if( (timeAgo < twoDays) && (timeAgo >= oneHour) ){
            var hours = ((timeAgo/1000)/60)/60;
            hours = Math.floor(hours);
            if( hours < 2 ) $t.html( 'an hour ago' );
                else $t.html( hours + ' hours ago' );
        } //else => do nothing (keep original content)
    });

});

=> http://jsfiddle.net/chgvcudL/

Upvotes: 0

comebal
comebal

Reputation: 946

Apparently, I have to modify the plugin. Have to set the "hours" to be less than 48 hours.

 var words = seconds < 45 && substitute($l.seconds, Math.round(seconds)) ||
 hours < 48 && substitute($l.hours, Math.round(hours)) ||

Upvotes: 1

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