Hello
This (=) is used in the variable references. in $ref=$array[2] line
$ref variable and $array[2] variable point at the same address. if you
assign any value to this any variables, both of them will change their
values because echo($ref) line display b on the screen.
--
Haydar TUNA
Peter Beckman wrote:
Because I'm trying to point out a problem with PHP, where setting a
reference when the other side is undefined or not set, PHP creates a
reference to a previously non-existent array item, just by setting a
reference. I don't think that should happen.
And? what's wrong
I'm looping through an array and I did this:
$rate = $mydata[$prefix];
Now, in some cases $mydata[$prefix] wasn't set/defined, so I expected $rate
to not be defined, or at least point to something that wasn't defined.
Instead, PHP 5.1.6 set $mydata[$prefix] to nothing.
If I had:
sry i don't get what u mean??
- Original Message -
From: Peter Beckman [EMAIL PROTECTED]
To: PHP-DB Mailing List php-db@lists.php.net
Sent: Tuesday, February 13, 2007 8:29 AM
Subject: [PHP-DB] Strange action with =
I'm looping through an array and I did this:
$rate = $mydata
On Tue, 13 Feb 2007, bedul wrote:
sry i don't get what u mean??
I'm looping through an array and I did this:
$rate = $mydata[$prefix];
This is how you assign a variable by reference. $rate should be a
reference to $mydata[$prefix], not a copy. If I change the value of
$rate, the
