GREAT.  This does the trick for me.  Thanks a lot.  I am now getting
the output I want, so I can now feel confident about inputting this into the
actual database.

-----Original Message-----
From: Katie Evans-Young [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 11:21 AM
To: NIPP, SCOTT V (SBCSI)
Subject: RE: [PHP-DB] Easy array question...


It's actually putting the whole of the array $other as the first element of
the array $other2. You're going to have to step through the array to copy
it.

foreach($other as $element) {
        $other2[] = $element;
} //end foreach

-----Original Message-----
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 11:10 AM
To: '[EMAIL PROTECTED]'; 'Hutchins, Richard'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Easy array question...


        Here is basically what I have and what I am getting...

        The section of code to copy and then process the copied array:

$other2 = array();
array_push($other2, $other);

foreach ($other2 as $test) {
        if ($test == "") {
                echo "NOTHING"."<br>";
        } else {
                echo $test."<br>";
        }
}

        Basically, what I want is if the array element is nothing, then
output as a test NOTHING.  If the array element has some value, then output
as a test that value.  Currently, from this code the output I am actually
getting is:

Array

        Thanks again.

-----Original Message-----
From: Jonathan Villa [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:58 AM
To: [EMAIL PROTECTED]; NIPP, SCOTT V (SBCSI); 'Hutchins, Richard';
[EMAIL PROTECTED]
Subject: RE: [PHP-DB] Easy array question...


Sorry, I just reread your post

$array_copy = array();
array_push($array_copy, $original_array);

echo $array_copy[0];//or whatever to test


---> Jonathan




-----Original Message-----
From: Jonathan Villa [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:56 AM
To: 'NIPP, SCOTT V (SBCSI)'; 'Hutchins, Richard'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Easy array question...

$array_copy = array();
array_push($array_copy, $_SESSION['original_array']);

echo $array_copy[0];//or whatever to test


---> Jonathan



-----Original Message-----
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:43 AM
To: 'Hutchins, Richard'; '[EMAIL PROTECTED]'
Subject: RE: [PHP-DB] Easy array question...

        I don't think that is the case because I am able to access the
original array in this same page without having to use the session
variable
reference.  Won't hurt to give it a try though.  Thanks for the
feedback.

-----Original Message-----
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:40 AM
To: '[EMAIL PROTECTED]'
Subject: RE: [PHP-DB] Easy array question...


Are you using:

$array2 = $_SESSION["array1"];

Not sure that will do the trick, but it was my first inclination.

Rich

> -----Original Message-----
> From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, March 12, 2003 11:37 AM
> To: '[EMAIL PROTECTED]'
> Subject: [PHP-DB] Easy array question...
>
>
>       I create an array on one page and then process it on
> the next page
> as I have registered the array as a session variable.  I need
> to basically
> process one of these arrays twice, so I want to make a copy
> of the array.
> The question is how do I do this?  I have tried the following:
>
> $array2 = $array1;
>
>       This doesn't seem to do the trick for me.  Is this
> correct and I am
> just missing something elsewhere, or is there another way to
> copy an array?
> Thanks.
>
> Scott Nipp
> Phone:  (214) 858-1289
> E-mail:  [EMAIL PROTECTED]
> Web:  http:\\ldsa.sbcld.sbc.com
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>

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