Question

I am using jquery tools date input. In my application i have a feature that allows the user to check his personal data based on date range.

I have an input text field which loads the date Input functionality. When the user select a date it displays it in : “dddd dd, mmmm yyyy” format in the text field.
What i am trying to do is to add 7 days (ie a week) and display the new date in the upper range container, to explain more this is how it looks:

From: Friday 18, May 2012
To: //here is where i want it to be : Friday 25, May 2012

I am trying to use the built in api function :addDay(amount) but it seems not to work.
I have the following code:

var ret_date = $(this).data("dateinput").getValue('dddd dd, mmmm yyyy');
//I tried: $(this).data("dateinput").addDay(7).getValue('dddd dd, mmmm yyyy')
//but it kept the same currently selected date
$("#to_field").html(ret_date);

I searched online and SO but couldn’t find a solution. I am not generally a front end developer so my knowledge with javascript/jquery is not advanced.

Any help is much appreciated

Answer 1

By the way, this is what i ended up doing and it worked:

$("#DateField").change(function(event, date) {
        //I used format: yyyy-mm-dd
        var ret_date= $(this).data("dateinput").getValue('yyyy-mm-dd');   
        //Format the date using javascript
        //--> For that 2 javascript prototypes have been created(check at the end of the answer)          
        var myDate = new Date(ret_date);
        myDate.setDate(myDate.getDate()+7);
        $("#FinalDateField").html(myDate.dateFormat("Y-m-d"));
});

The prototypes:

Date.prototype.dateFormat = function(format) {
  var result = "";
  for (var i = 0; i < format.length; ++i) {
    result += this.dateToString(format.charAt(i));
  }
  return result;
}

Date.prototype.dateToString = function(char) {
  switch (char) {
  case "Y":
    return this.getFullYear();
  case "m":
    return this.getMonth() + 1; //Since months are zero based, add 1
  case "d":
    return this.getDate();
  default:
    return char;
 }
}