Date from $ help please

cheese

Good evening all,

I have R'd T*M but can't find how to do this - I'm sure it is simple;

I have a variable called $col1new which is 31-12-2004

I would like to covert this so that the variable is 31st December 2004

Please could someone assist an oldy....

Thanks,

Neil.

Davy

Could you not have just called the date() or now() to the variable and formatted it accordingly?

Anyway I'm assuming you already have this date (and proberly a good few others) and all you want to do is convert them.

Well, here at PHPBuilder we don't like giving the answer away freely in detail. We like to either give a small snippet of example code or point you in the right direction.

Ok here goes...
It would be possible to just write code to convert JUST these numbers to what you need but you probably want to input any date of "dd-mm-yyyy" format to convert it to "dd/p month yyyy".
please note these letters areb't the proper date formats, just examples

So, you want to first divide the date up into seperate variables, run IF statements on these numbers to convert them into the desired format and code some sort of output for these dates.

Ok sod it, I'll do some of the code but it won't be completely clean. It won't return any 'or die's or anything like that.

$NewDateArray = explode("-","$col1new");
// This will take the date and divide it up. The explode function cuts where the -s are and puts the values into the array called $NewDateArray.

/*************************
 Day number conversion below
*************************/
if ($NewDateArray[0] == 31)
  {
  $day = "31st ";
  }
/* Repeat this IF statement for all 31 days. You can copy/paste/modify that code yourself. This is a small snippet :) */

/**************************
  Month number conversion below
**************************/
if ($NewDateArray[1] == 12)
  {
  $month = "December ";
  }
/* Repeat this one, too */

/****************************
  Year number conversion below
****************************/
//No actual conversion as such here, we just need to take it out of the array for easy handling later when we merge all these variables into one for the complete date.

$year = $NewDateArray[2];

/****************************
  Now we put all of these together 
  to form the complete string 
****************************/

$NewDate = $day . $month . $year;

As you would expect, this code could be done differently.
How is the date saved to this variable, anyway? Does is take the current date and save that directly, or does it take a user's input?

Anyway this should help. Hope it does!

bubblenut

I would use [man]explode()[/man], [man]list()/man, [man]mktime()[/man] and [man]date()[/man] like this

list($day, $month, $year) = explode('-', $col1new);
echo date('jS F Y', mktime(0, 0, 0, $month, $day, $year));

HTH
Bubble

Installer

This is the answer I came up with to give away:

function d_m_Y_to_jS_F_Y($date, $separator = '-')
{
    $date_arr = explode($separator, $date);
    $rev_date = $date_arr[2] . '-' . $date_arr[1] . '-' . $date_arr[0];
    $date_time = strtotime($rev_date);
    $new_date = date('jS F Y', $date_time);
    return $new_date;
}

But bubblenut's is better, so I took the liberty of putting it into a function:

function convert_date($in_date, $separator = '-')
{
    list($day, $month, $year) = explode($separator, $in_date); 
    return date('jS F Y', mktime(0, 0, 0, $month, $day, $year));
}

cheese

Thanks for the replies.....

For reference this is what I had written. It works, but I just feel it is a little OTT for what it does.

For the days.....

$col1new1 = substr($col1new ,0,2);
 if (($col1new1 == '01') OR ($col1new1 == '21') OR ($col1new1 == '31'))
 	{
 	$col1new1 = $col1new1."st";
  } 
 if (($col1new1 == '02') OR ($col1new1 == '22'))
  {
 	$col1new1 = $col1new1."nd";
	} 
 if (($col1new1 == '03') OR ($col1new1 == '23'))
  {
 	$col1new1 = $col1new1."rd";
	} 
$temp = substr($col1new1,0,2);
 if (($temp != '01') AND ($temp != '02') AND ($temp != '03') AND ($temp != '21') AND ($temp != '22') AND ($temp != '23') AND ($temp != '31'))
  {
	$col1new1 = $col1new1."th";
  }
 if (substr($col1new1,0,1) == '0')
  { 
  $col1new1 = substr($col1new1,1,3);
  }

For the months.... (more or less the same as Davy)

$col1new2 = substr($col1new,3,2);
 if ($col1new2 == 1)
  {
  $col1new2 = 'January';
  }
// Repeated 12 times.......

And for the year and putting it back together again.....

$col1new3 = substr($col1new ,6,4);

$col1new = $col1new1."&nbsp".$col1new2."&nbsp".$col1new3;

Still think it looks very messy.... but hey, it works.

Thanks again,

Neil.

Installer

So...what are you using now?

examancer

maybe i'm missing something, but wouldn't this be tons easier?:

$newdate = date("jS F Y", strtotime($col1new));

strtotime will take any valid date representation and make it a time stamp... then all you have to do is use date() to format the timestamp to your "DDth Month YYYY" format.

Installer

A "valid date representation" is any representation the function will take. "31-12-2004" isn't one of them. The ones it will take are given here. When you try this:

$col1new = '31-12-2004';
$newdate = date("jS F Y", strtotime($col1new));
echo $newdate;

you'll see it doesn't work.

examancer

ah, so you used explode() in your example to put the month before the days so that it will be accepted by strtotime(). didn't catch that.

cheese

Thanks to all for their advice.....

My son looked at my code, laughed.... tutted.... laughed a bit more and then did it in Installers format......

This is what comes with the problem of trying to work in PHP on very rare accassions, you tend to forget what you have picked up.... that mixed with the onset of old age..

Thanks again all,

Neil.

Installer

I'm glad we were able to help you out, cheese.

The posts by examancer made me think about what a poor job of commenting code I (and a lot of us) do. Or, what a great job of not commenting code. It'd probably be a lot more helpful to people, especially but not only those new to PHP, if they could more easily see what is being done and why. I'll try to improve.