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From: Olavi Ivask on 25 Jan 2010 14:44 Don't you like my solution? Regards, Olavi On Jan 25, 2010, at 9:39 PM, Karl DeSaulniers wrote: > Hi, > $req_user_level is a MySql return result and I am trying to utilize > shortened code to evaluate whither $req_user_level == 0 || 1 || 2 || > 3 || 4 || 5 || 6 || 7 || 8 || 9, if it matches a number, return the > text associated with that number. > "Guest" || "Regular User" || "Intl. User" || "Contractor" || > "Employee" || "Sales" || "Investor" || "Human Resources" || > "Administrator"; > > Karl > > > On Jan 25, 2010, at 1:31 PM, Olavi Ivask wrote: > >> Hi, >> >> did you mean something like this? >> >> $level_names = array("Guest", "Regular User", "Intl. User", >> "Contractor", "Employee", >> "Sales", "Investor", "Human Resources", "Administrator"); >> >> $user_level = $level_names[$req_user_level]; >> >> Regards, >> Olavi Ivask >> >> >> On Jan 25, 2010, at 9:13 PM, Karl DeSaulniers wrote: >> >>> Hello List, >>> Trying to learn the right way to code this line. >>> Can anyone tell me if I am doing this the right way? >>> >>> if $req_user_level == 0 || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || >>> 9 ? "Guest" || "Regular User" || "Intl. User" || "Contractor" || >>> "Employee" || "Sales" || "Investor" || "Human Resources" || >>> "Administrator"; >>> >>> I am wanting to stray from the if(foo == bar) { routine. ;) >>> >>> Thanks, >>> >>> Karl DeSaulniers >>> Design Drumm >>> http://designdrumm.com >>> >> > > Karl DeSaulniers > Design Drumm > http://designdrumm.com >
From: Karl DeSaulniers on 25 Jan 2010 14:58 Thank you for this as well. Question? What part is "in_array" playing? Is it comparing $req_user_level to array()? Because the text "Guest", etc.. is not in $req_user_level on the database. In other words, is it checking the value of $req_user_level to see if "Guest" is in it? Karl On Jan 25, 2010, at 1:32 PM, Peter Beckman wrote: > On Mon, 25 Jan 2010, Karl DeSaulniers wrote: > >> Hello List, >> Trying to learn the right way to code this line. >> Can anyone tell me if I am doing this the right way? >> >> if $req_user_level == 0 || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || >> 9 ? "Guest" || "Regular User" || "Intl. User" || "Contractor" || >> "Employee" || "Sales" || "Investor" || "Human Resources" || >> "Administrator"; >> >> I am wanting to stray from the if(foo == bar) { routine. ;) > > if (in_array($req_user_level, array > (0,1,2,3,4,5,6,7,8,9,"Guest",...))) { > // do stuff > } > > ---------------------------------------------------------------------- > ----- > Peter Beckman > Internet Guy > beckman(a)angryox.com http:// > www.angryox.com/ > ---------------------------------------------------------------------- > ----- > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > Karl DeSaulniers Design Drumm http://designdrumm.com
From: Peter Beckman on 25 Jan 2010 15:24 On Mon, 25 Jan 2010, Karl DeSaulniers wrote: > Thank you for this as well. > Question? What part is "in_array" playing? > Is it comparing $req_user_level to array()? > Because the text "Guest", etc.. is not in $req_user_level on the database. > In other words, is it checking the value of $req_user_level to see if "Guest" > is in it? Sorry, I missed the question mark. in_array isn't appropriate here. The previous poster has it right, assuming $req_user_level is an integer of 0..9. $levels = array("Guest", "Regular User", 'Intl. User', ...) // the array is 0 , 1 , 2 , ...) Going a little further: if (!empty($levels[$req_user_level])) { // is both set and doesn't evaluate to false echo "The user is a {$levels[$req_user_level]}.\n"; } else { // The $req_user_level was not a valid level. echo "The returned req_user_level was not valid.\n"; } Which would output, if $req_user_level was 1 (one): The user is a Regular User. Then you know you have a valid user level. Careful though -- sometimes 0 will be returned on a failure, depending on your SQL. Beckman --------------------------------------------------------------------------- Peter Beckman Internet Guy beckman(a)angryox.com http://www.angryox.com/ ---------------------------------------------------------------------------
From: Karl DeSaulniers on 25 Jan 2010 15:33
Yes, $req_user_level is an int between 0 and 9. Yes, the other code worked great. Thank you for your response though. That is definitely a good way to cross check. Thanks again for your responses. Best, Karl On Jan 25, 2010, at 2:24 PM, Peter Beckman wrote: > On Mon, 25 Jan 2010, Karl DeSaulniers wrote: > >> Thank you for this as well. >> Question? What part is "in_array" playing? >> Is it comparing $req_user_level to array()? >> Because the text "Guest", etc.. is not in $req_user_level on the >> database. >> In other words, is it checking the value of $req_user_level to see >> if "Guest" is in it? > > Sorry, I missed the question mark. in_array isn't appropriate > here. The > previous poster has it right, assuming $req_user_level is an > integer of > 0..9. > > $levels = array("Guest", "Regular User", 'Intl. User', ...) > // the array is 0 , 1 , 2 , ...) > > Going a little further: > > if (!empty($levels[$req_user_level])) { // is both set and > doesn't evaluate to false > echo "The user is a {$levels[$req_user_level]}.\n"; > } else { > // The $req_user_level was not a valid level. > echo "The returned req_user_level was not valid.\n"; > } > > Which would output, if $req_user_level was 1 (one): > > The user is a Regular User. > > Then you know you have a valid user level. Careful though -- > sometimes 0 > will be returned on a failure, depending on your SQL. > > Beckman > ---------------------------------------------------------------------- > ----- > Peter Beckman > Internet Guy > beckman(a)angryox.com http:// > www.angryox.com/ > ---------------------------------------------------------------------- > ----- > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > Karl DeSaulniers Design Drumm http://designdrumm.com |