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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Easy array question...
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.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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Easy array question...
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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php --- Incoming mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.445 / Virus Database: 250 - Release Date: 1/21/2003 --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.445 / Virus Database: 250 - Release Date: 1/21/2003 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php