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Re: [PHP] Problem with array

Posted by Rory Browne on 06/17/05 09:34

Not sure if it works for numeric indices, but maybe you could replace
$piece[3] with (array_key_exists(3, $piece) ? $piece[3] : ""). If you
want you could abstract that into a function, like

function array_access_element($key, $srch_array, $def=""){
return array_key_exists($key, $srch_array) ? $srch_array[$key] : $def;
}


On 6/17/05, Ross <ross@aztechost.com> wrote:
> As with my previous post the problem is the pieces of the array can vary
> from 1 to 4 items. So pieces 3 and 4 are often undefined giving the
> 'undefined index' notice. All I really want to do is display the array
> pieces if they EXIST. But as they are inside a echo statement so I can't
> even to a for loop...can I?
>
>
> Any ideas?
>
> R.
>
>
> if ($quantity == 0){
>
> }
> else {
>
> $pieces = explode(" ", $quantity);
>
>
> $formatted_price = sprintf('%0.2f', $pricecode);
> echo "<table width=\"240\" border=\"0\" cellpadding=\"2\"
> cellspacing=\"5\"><tr><td valign=\"top\" align=\"right\"
> width=\"40\">$pieces[0]</td><td align=\"left\" width=\"60\"></td><td
> align=\"left\" width=\"200\">$pieces[1] $pieces[2] $pieces[3]
> $pieces[4]</td><td valign=\"top\" align=\"left\"
> width=\"80\">$formatted_price</td></tr></table>";
>
>
>
>
>
> }
> }
>
> --
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>

 

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