RE: [PHP-DB] Easy array question...

2003-03-12 Thread NIPP, SCOTT V (SBCSI) 
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"."";
} else {
echo $test."";
}
}

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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RE: [PHP-DB] Easy array question...

2003-03-12 Thread Jonathan Villa 
Try it with the original array first, just for testing

 
---> Jonathan
 
 
 

-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"."";
} else {
echo $test."";
}
}

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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RE: [PHP-DB] Easy array question...

2003-03-12 Thread NIPP, SCOTT V (SBCSI) 
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"."";
} else {
echo $test."";
}
}

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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RE: [PHP-DB] Easy array question...

2003-03-12 Thread Jonathan Villa 
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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RE: [PHP-DB] Easy array question...

2003-03-12 Thread Jonathan Villa 
$array_copy = array();
array_push($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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RE: [PHP-DB] Easy array question...

2003-03-12 Thread NIPP, SCOTT V (SBCSI) 
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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RE: [PHP-DB] Easy array question...

2003-03-12 Thread Hutchins, Richard 
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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