beginner problem - using date("M", $i) where $i is an integer

[bissatch@yahoo.co.uk]

Hi,

I am trying to use the following function:


echo date("M", $i)


to echo the date that the value of a variable, $i, represents (eg. 1 =
Jan, 5 = May, 11 = Nov). Unfortunetely, it returns only Jan for any
number which I assume is the 1970 date that this function returns when
you screw up. I know that when I have the string value of '2005-05-01'
in a variable, I have to convert it in the following function:


echo date("M", strtotime($strd ate))


Is something similar required when I use an integer? Thanks

Burnsy

[Ken Robinson]

Re: beginner problem - using date("M&qu ot;, $i) where $i is an integer



bissatch@yahoo. co.uk wrote:[color=blue]
> Hi,
>
> I am trying to use the following function:
>
>
> echo date("M", $i)
>
>
> to echo the date that the value of a variable, $i, represents (eg. 1 =
> Jan, 5 = May, 11 = Nov). Unfortunetely, it returns only Jan for any
> number which I assume is the 1970 date that this function returns when
> you screw up. I know that when I have the string value of '2005-05-01'
> in a variable, I have to convert it in the following function:[/color]

The second parameter to the date() function is the number of seconds
since 1970-01-01 (the UNIX beginning of time). To do what you want to
do, use

echo date("M",strtot ime('2005-'.$i.'-01'));

Ken

[Chung Leong]

Re: beginner problem - using date("M&qu ot;, $i) where $i is an integer

This is slightly more efficient, I think:

echo date("M", mktime(0, 0, 0, $i));