On Tue, 2010-10-05 at 15:46 -0400, Steve Staples wrote:
> On Tue, 2010-10-05 at 20:35 +0100, Ashley Sheridan wrote:
> > On Tue, 2010-10-05 at 15:28 -0400, chris h wrote:
> >
> > > Benchmark and find out! :)
> > >
> > > What are you using this for? Unless you are doing something crazy it
> > > probably doesn't matter, and you should pick whichever you feel looks
> > > nicer
> > > / is easier to code in / etc.
> > >
> > > Chris H.
> > >
> > > On Tue, Oct 5, 2010 at 3:23 PM, saeed ahmed <saeed....@gmail.com> wrote:
> > >
> > > > $a = 'hey';
> > > > $b = 'done';
> > > >
> > > > $c = $a.$b;
> > > > $c = "$a$b";
> > > >
> > > > which one is faster for echo $c.
> > > >
> >
> >
> > As far as I'm aware, the first of the two will be faster, but only just.
> > As Saeed mentioned, the difference will be negligible, and unless you
> > plan to run a line like that in a loop or something hundreds of
> > thousands of times, you probably won't notice any difference.
> > Thanks,
> > Ash
> > http://www.ashleysheridan.co.uk
> >
> >
>
>
> to be proper, shouldn't it technically be
> $c = "{$a}{$b}";
>
> ??
>
> Steve.
>
>
It doesn't have to use the braces. The braces only tell PHP exactly
where to stop parsing the current variable name. The following examples
wouldn't work without them:
$var = 'hello ';
$arr = array('msg 1'=>'hello','msg 2'=>'world');
echo "{$var}world";
echo "{$arr['msg 1']}{$arr['msg 2']}";
Without the braces, in the first example PHP would look for a variable
called $varworld, and in the second it would be looking for a simple
scaler called $arr, not the array value you wanted.
Thanks,
Ash
http://www.ashleysheridan.co.uk
