Question

With this line of code

$("body").html($("body").html().replace(/1800/g,'6:00pm'));

I can replace the word, but how can I make it more dynamic with this script?

var getFormattedTime = function (fourDigitTime){
    var hours24 = parseInt(fourDigitTime.substring(0,2));
    var hours = ((hours24 + 11) % 12) + 1;
    var amPm = hours24 > 11 ? 'pm' : 'am';
    var minutes = fourDigitTime.substring(2);

    return hours + ':' + minutes + amPm;
};

My HTML code is e.g

Your class will start at 1900, please be there by 1845.

Which will be counted to

Your class will start at 07:00pm, please be there by 06:45pm.

I tried this:

$("body").html($("body").html().replace(fourDigitTime, getFormattedTime(fourDigitTime)));

var fourDigitTime = document.getElementById('fourDigitTime').value;

var getFormattedTime = function (fourDigitTime){
var hours24 = parseInt(fourDigitTime.substring(0,2));
var hours = ((hours24 + 11) % 12) + 1;
var amPm = hours24 > 11 ? 'pm' : 'am';
var minutes = fourDigitTime.substring(2);

return hours + ':' + minutes + amPm;
};

It does not work..

Answer 1

[edited after EmersonF cleared up that all four letter times should be replaced]

You can use a function as the second argument to replace:

function formatTime(str, p1) {
   // do something with p1
   return // the formated time
}

$("body").html($("body").html().replace(/(\d?\d\d\d)/g, formatTime));

The part of the string that is “captured” using the parenthesis in the
regular expression is handed to the function as the second argument (here called p1)

See http://jsfiddle.net/bjelline/pb8d2/ for a working example

Note: the regular expression cannot tell the year 2015 apart from quater
past 8pm, and will convert the year to a time!