NYCPHP Meetup

[Analysis & Solutions]

On Fri, May 30, 2003 at 11:52:06AM -0400, soazine at pop.erols.com wrote:
> Ok I will do the best I can, although it won't make the problem clearer (at
> least to me):

It absolutely helps.


> Warning: Invalid argument supplied for foreach() in
> /users/ppowell/web/profile/display.php on line 92
> 
> Line 92: 
> foreach ($browser as $key => $val) 
>      echo "$key = $val<br>";

I'll bet $browser isn't an array.


> Furthermore I do not understand
> 
> $browserArray = array($key, $val);

Dude, did you write the exact syntax I wrote?  No.

The syntax you wrote makes an enumerated array with two values.  What I 
think you really want to be doing is creating an associative array with 
one key value pair.  Right?  So, use "$key => $val" instead of "$key, 
$val"

But, since you're pulling stuff from the $browser array already, why are 
you creating a second array with the same exact data?  I suspect you're 
off course...


> I have never seen that before

The manual is an excellent resource.  Have you looked at the array 
section?  Finding it is as simple as typing "php.net/array" into most web 
browsers.

--Dan

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