> -----Original Message----- > From: Jigar Dhulla [mailto:jigar.tidus@xxxxxxxxx] > Sent: Wednesday, July 22, 2015 10:26 PM > To: hadi > Cc: php-general@xxxxxxxxxxxxx General > Subject: Re: asterisk with php > > On Wed, Jul 22, 2015 at 11:51 PM, hadi <almarzuki2011@xxxxxxxxxxx> wrote: > > > Hi, > > > > I have project, Asterisk with phpagi, when I pass variable to function > > I get error "Notice: Undefined variable: pin". > > > > Here is my code > > > > > > > > #!/usr/local/bin/php -q > > <?php > > > > error_reporting(E_ALL); > > ini_set('display_errors', 1); > > > > require("phpagi.php"); > > require("database.php"); > > > > > > > > > > > > $agi=new AGI(); > > > > > > > > $agi->answer(); > > > > > > for ($x = 0; $x <= 3; $x++) > > { > > > > > > > > > > $result = $agi->get_data('/var/lib/asterisk/agi-bin/ivr-sound1', 3000, > > 20); > > > > > > $pass = $result['result']; > > > > > > $query = $conn->prepare("SELECT COUNT(*) FROM userinfo where > > pin=:pin"); $query->bindParam(':pin', $pass); > > > > $query->execute(); > > > > $num_rows = $query->fetchColumn(); > > > > if($num_rows == 1) > > > > > > > > > > { > > $agi->text2wav("thankyou"); > > > > > > > > $agi->text2wav("please enter number to dial for more information press > > zero"); > > > > > > $num = $agi->exec("Read","pin,,4,3,120"); > > > > $temp = $agi->get_variable("pin"); > > > > $pin = trim($temp["data"]); > > > > global $pin; > > > > break; > > > > } > > > > > > if ($x == 3) > > > > { > > $agi->hangup(); > > > > } > > > > } > > > > > > > > > > dialout($pin); > > > > > > function > > > > dialout($pin) > > > > { > > > > > > > > $agi=new AGI(); > > > > $agi->exec_dial("SIP/$pin"); > > > > $Answeredtime = $agi->get_variable ("ANSWEREDTIME"); > > > > $agi->hangup (); > > > > > > } > > > > > > > > > > ?> > > > > > > > > > > -- > > PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: > > http://www.php.net/unsub.php > > > > > Hi, > > That "Notice" means you are trying to use/access that variable even though it > is not yet defined/set. > > Looking at your code, if I am not wrong *$pin* is defined only when > *$num_rows == 1, *right ? Hence, when > *dialout($pin) *is called *$pin *will remain undefined in other cases, so the > notice(this is just one of the possibility). > > So the best solution is to use *isset($pin) *where you think *$pin *can be > not defined/set. > > > *if(isset($pin) && !empty($pin)){* > > * dialout($pin);* > *}* > > Also in production/live environment, don't forget to suppress the notice > *error_reporting(E_ALL & ~E_NOTICE & ~E_WARNING & ~E_STRICT & > ~E_DEPRECATED); *you might not want your users to see those notice or > warning. Avoid suppressing such things in development environment. > Looking at your code, if I am not wrong *$pin* is defined only when > *$num_rows == 1, *right ? Yes your right.. When I do "var_dump($pin)" at the beginning of the function it show the variable correct. But when I do "var_dump($pin)" after "$agi=new AGI();" it show nothing. I think "$agi=new AGI();" it wipe out my "$pin" variable.... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
