calc field php syntax - an easy php problem

[jcuming]

Hi. Can anyone help? for someone who knows php, this is probably a no-brainer. I don't.

I've a list with several fields. One of the fields is called 'assistant___Role' and is a date field.
Another field is called 'assistant___flag' and is a calc field.
I want the 'flag' field to be either null, or else to show the value of the 'Role' field.
I've put the code below in the calculation box for 'flag'. The code sort of works. But the output is in the format
Y-m-d h:s etc) I want it in the format 'd/m/y' (and no hours/seconds).


$thresholddate = strtotime("-36 Months");
$roledays = strtotime('{assistant___Role}');
$roledate = '{assistant___Role}';

if ($roledays < $thresholddate )
{ return date( $roledate) ; }

I would guess to get the format I want, I'd just change the line:

$roledate = '{assistant___Role}';
to
$roledate = date("d-m-Y",'{assistant___Role}');
But when I do, all output dates become 01-01-1970.
What am I doing wrong? I have no knowledge of php, so have just cobbled thi together from various websites.
THanks
James

 

[cheesegrits]

Try ...

$roledate = date("d-m-Y", $roledays);

The date() function expects a UNIX timestamp, which you already have in $roledays (that's what strtotime() does). Your existing code is trying to feed an already formatted time string to date().

-- hugh

 

[jcuming]

Thanks so much cheese. Its dangerous when people like me try php. Will try it and let you know.

 

[cheesegrits]

I'm just as dangerous. The more you know, the worse the damage you can do, more quickly.

-- hugh

 

[jcuming]

You beauty! It WORKS! Ha!
Thanks v much again
Jim