call_user_func_array throws an error when the method exist












1














This is how I am reaching the _call method:



$model->delivery_price = $currencyConverter->convertPriceByGivenCurrencies(

                        $model->delivery_price,

                        $currency->id,

                        $model->order_currency

                    );


The function throws an error but the method exists below it. My __call looks like:



public function __call($name, $params)

    {

        if(method_exists(CurrencyConverter::className(), $name)){

            if($params[0] == 0 || $params[0]){

                call_user_func_array($name, $params);

            }else{

                throw new Exception('Price must be a valid number!');

            }

        }

        throw new NotFoundException('Function doesn't exist');

    }


It passes the if condition but after that the error occures:



call_user_func_array() expects parameter 1 to be a valid callback, function 'convertPriceByGivenCurrencies' not found or invalid function name


And this is the convertPriceByGivenCurrencies method which is landed in the below the _call:



protected function convertPriceByGivenCurrencies($product_price, $product_price_currency_id, $select_currency_id)

    {

      ............

    }


What am I doing wrong here ? Thank you!






php







share|improve this question






















  • When you are checking whether the method name exists, you are passing a class name to do so. But where in your call_user_func_array call are you referring to that same class …? You are passing the function name only there, so it looks for a standalone function of that name.
    – misorude
    Nov 13 '18 at 7:44
















1














This is how I am reaching the _call method:



$model->delivery_price = $currencyConverter->convertPriceByGivenCurrencies(

                        $model->delivery_price,

                        $currency->id,

                        $model->order_currency

                    );


The function throws an error but the method exists below it. My __call looks like:



public function __call($name, $params)

    {

        if(method_exists(CurrencyConverter::className(), $name)){

            if($params[0] == 0 || $params[0]){

                call_user_func_array($name, $params);

            }else{

                throw new Exception('Price must be a valid number!');

            }

        }

        throw new NotFoundException('Function doesn't exist');

    }


It passes the if condition but after that the error occures:



call_user_func_array() expects parameter 1 to be a valid callback, function 'convertPriceByGivenCurrencies' not found or invalid function name


And this is the convertPriceByGivenCurrencies method which is landed in the below the _call:



protected function convertPriceByGivenCurrencies($product_price, $product_price_currency_id, $select_currency_id)

    {

      ............

    }


What am I doing wrong here ? Thank you!






php







share|improve this question






















  • When you are checking whether the method name exists, you are passing a class name to do so. But where in your call_user_func_array call are you referring to that same class …? You are passing the function name only there, so it looks for a standalone function of that name.
    – misorude
    Nov 13 '18 at 7:44














1












1








1







This is how I am reaching the _call method:



$model->delivery_price = $currencyConverter->convertPriceByGivenCurrencies(

                        $model->delivery_price,

                        $currency->id,

                        $model->order_currency

                    );


The function throws an error but the method exists below it. My __call looks like:



public function __call($name, $params)

    {

        if(method_exists(CurrencyConverter::className(), $name)){

            if($params[0] == 0 || $params[0]){

                call_user_func_array($name, $params);

            }else{

                throw new Exception('Price must be a valid number!');

            }

        }

        throw new NotFoundException('Function doesn't exist');

    }


It passes the if condition but after that the error occures:



call_user_func_array() expects parameter 1 to be a valid callback, function 'convertPriceByGivenCurrencies' not found or invalid function name


And this is the convertPriceByGivenCurrencies method which is landed in the below the _call:



protected function convertPriceByGivenCurrencies($product_price, $product_price_currency_id, $select_currency_id)

    {

      ............

    }


What am I doing wrong here ? Thank you!






php







share|improve this question













This is how I am reaching the _call method:



$model->delivery_price = $currencyConverter->convertPriceByGivenCurrencies(

                        $model->delivery_price,

                        $currency->id,

                        $model->order_currency

                    );


The function throws an error but the method exists below it. My __call looks like:



public function __call($name, $params)

    {

        if(method_exists(CurrencyConverter::className(), $name)){

            if($params[0] == 0 || $params[0]){

                call_user_func_array($name, $params);

            }else{

                throw new Exception('Price must be a valid number!');

            }

        }

        throw new NotFoundException('Function doesn't exist');

    }


It passes the if condition but after that the error occures:



call_user_func_array() expects parameter 1 to be a valid callback, function 'convertPriceByGivenCurrencies' not found or invalid function name


And this is the convertPriceByGivenCurrencies method which is landed in the below the _call:



protected function convertPriceByGivenCurrencies($product_price, $product_price_currency_id, $select_currency_id)

    {

      ............

    }


What am I doing wrong here ? Thank you!






php




php






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 13 '18 at 7:38









Toma TomovToma Tomov

612216




612216












  • When you are checking whether the method name exists, you are passing a class name to do so. But where in your call_user_func_array call are you referring to that same class …? You are passing the function name only there, so it looks for a standalone function of that name.
    – misorude
    Nov 13 '18 at 7:44


















  • When you are checking whether the method name exists, you are passing a class name to do so. But where in your call_user_func_array call are you referring to that same class …? You are passing the function name only there, so it looks for a standalone function of that name.
    – misorude
    Nov 13 '18 at 7:44
















When you are checking whether the method name exists, you are passing a class name to do so. But where in your call_user_func_array call are you referring to that same class …? You are passing the function name only there, so it looks for a standalone function of that name.
– misorude
Nov 13 '18 at 7:44




When you are checking whether the method name exists, you are passing a class name to do so. But where in your call_user_func_array call are you referring to that same class …? You are passing the function name only there, so it looks for a standalone function of that name.
– misorude
Nov 13 '18 at 7:44












2 Answers
2






active

oldest

votes


















1














$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.



So to call it, assuming it is a static method, you would need something like:



CurrencyConverter::$name(...$params);


Note that you need the ... operator to unpack $params






share|improve this answer























  • But the unpacking is avaible in 5.6 only. At least this is what the PHPStorm says :)
    – Toma Tomov
    Nov 13 '18 at 8:04










  • @TomaTomov That's right, if you are on an earlier version you should probably upgrade :-)
    – jeroen
    Nov 13 '18 at 8:09










  • In fact I am on PHP Version 7.2.6-1. it is not usable on higher version ?
    – Toma Tomov
    Nov 13 '18 at 8:12












  • @TomaTomov It is available from 5.6 on, so you can use it without any problems. Does PHPStorm know you are on 7.2?
    – jeroen
    Nov 13 '18 at 8:28








  • 1




    Guess he doesn't know because it is working :) Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:41



















1














Calling it with



call_user_func_array($name, $params);


it is expecting a standalone function called $name.



As it's a method in a class you need to add this information to the callable, if you want to call it on the current instance then use 



call_user_func_array(array($this,$name), $params);


If it isn't a method in the current instance, then replace $this with the appropriate instance. Or change the method to be static and replace $this with the class name.






share|improve this answer























  • __call and protected function convertPriceByGivenCurrencies are both in CurrencyConverter class. And the variable $currencyConverter is instance of the CurrencyConverter class. Sorry about the unclear explanation :)
    – Toma Tomov
    Nov 13 '18 at 8:03












  • Your answer also works and both gave me some new knowledges. I wish I could mark more than 1 best answers. Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:51











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.



So to call it, assuming it is a static method, you would need something like:



CurrencyConverter::$name(...$params);


Note that you need the ... operator to unpack $params






share|improve this answer























  • But the unpacking is avaible in 5.6 only. At least this is what the PHPStorm says :)
    – Toma Tomov
    Nov 13 '18 at 8:04










  • @TomaTomov That's right, if you are on an earlier version you should probably upgrade :-)
    – jeroen
    Nov 13 '18 at 8:09










  • In fact I am on PHP Version 7.2.6-1. it is not usable on higher version ?
    – Toma Tomov
    Nov 13 '18 at 8:12












  • @TomaTomov It is available from 5.6 on, so you can use it without any problems. Does PHPStorm know you are on 7.2?
    – jeroen
    Nov 13 '18 at 8:28








  • 1




    Guess he doesn't know because it is working :) Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:41
















1














$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.



So to call it, assuming it is a static method, you would need something like:



CurrencyConverter::$name(...$params);


Note that you need the ... operator to unpack $params






share|improve this answer























  • But the unpacking is avaible in 5.6 only. At least this is what the PHPStorm says :)
    – Toma Tomov
    Nov 13 '18 at 8:04










  • @TomaTomov That's right, if you are on an earlier version you should probably upgrade :-)
    – jeroen
    Nov 13 '18 at 8:09










  • In fact I am on PHP Version 7.2.6-1. it is not usable on higher version ?
    – Toma Tomov
    Nov 13 '18 at 8:12












  • @TomaTomov It is available from 5.6 on, so you can use it without any problems. Does PHPStorm know you are on 7.2?
    – jeroen
    Nov 13 '18 at 8:28








  • 1




    Guess he doesn't know because it is working :) Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:41














1












1








1






$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.



So to call it, assuming it is a static method, you would need something like:



CurrencyConverter::$name(...$params);


Note that you need the ... operator to unpack $params






share|improve this answer














$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.



So to call it, assuming it is a static method, you would need something like:



CurrencyConverter::$name(...$params);


Note that you need the ... operator to unpack $params







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 13 '18 at 7:49

























answered Nov 13 '18 at 7:44









jeroenjeroen

81.8k1698122




81.8k1698122












  • But the unpacking is avaible in 5.6 only. At least this is what the PHPStorm says :)
    – Toma Tomov
    Nov 13 '18 at 8:04










  • @TomaTomov That's right, if you are on an earlier version you should probably upgrade :-)
    – jeroen
    Nov 13 '18 at 8:09










  • In fact I am on PHP Version 7.2.6-1. it is not usable on higher version ?
    – Toma Tomov
    Nov 13 '18 at 8:12












  • @TomaTomov It is available from 5.6 on, so you can use it without any problems. Does PHPStorm know you are on 7.2?
    – jeroen
    Nov 13 '18 at 8:28








  • 1




    Guess he doesn't know because it is working :) Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:41


















  • But the unpacking is avaible in 5.6 only. At least this is what the PHPStorm says :)
    – Toma Tomov
    Nov 13 '18 at 8:04










  • @TomaTomov That's right, if you are on an earlier version you should probably upgrade :-)
    – jeroen
    Nov 13 '18 at 8:09










  • In fact I am on PHP Version 7.2.6-1. it is not usable on higher version ?
    – Toma Tomov
    Nov 13 '18 at 8:12












  • @TomaTomov It is available from 5.6 on, so you can use it without any problems. Does PHPStorm know you are on 7.2?
    – jeroen
    Nov 13 '18 at 8:28








  • 1




    Guess he doesn't know because it is working :) Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:41
















But the unpacking is avaible in 5.6 only. At least this is what the PHPStorm says :)
– Toma Tomov
Nov 13 '18 at 8:04




But the unpacking is avaible in 5.6 only. At least this is what the PHPStorm says :)
– Toma Tomov
Nov 13 '18 at 8:04












@TomaTomov That's right, if you are on an earlier version you should probably upgrade :-)
– jeroen
Nov 13 '18 at 8:09




@TomaTomov That's right, if you are on an earlier version you should probably upgrade :-)
– jeroen
Nov 13 '18 at 8:09












In fact I am on PHP Version 7.2.6-1. it is not usable on higher version ?
– Toma Tomov
Nov 13 '18 at 8:12






In fact I am on PHP Version 7.2.6-1. it is not usable on higher version ?
– Toma Tomov
Nov 13 '18 at 8:12














@TomaTomov It is available from 5.6 on, so you can use it without any problems. Does PHPStorm know you are on 7.2?
– jeroen
Nov 13 '18 at 8:28






@TomaTomov It is available from 5.6 on, so you can use it without any problems. Does PHPStorm know you are on 7.2?
– jeroen
Nov 13 '18 at 8:28






1




1




Guess he doesn't know because it is working :) Thank you very much!
– Toma Tomov
Nov 13 '18 at 8:41




Guess he doesn't know because it is working :) Thank you very much!
– Toma Tomov
Nov 13 '18 at 8:41













1














Calling it with



call_user_func_array($name, $params);


it is expecting a standalone function called $name.



As it's a method in a class you need to add this information to the callable, if you want to call it on the current instance then use 



call_user_func_array(array($this,$name), $params);


If it isn't a method in the current instance, then replace $this with the appropriate instance. Or change the method to be static and replace $this with the class name.






share|improve this answer























  • __call and protected function convertPriceByGivenCurrencies are both in CurrencyConverter class. And the variable $currencyConverter is instance of the CurrencyConverter class. Sorry about the unclear explanation :)
    – Toma Tomov
    Nov 13 '18 at 8:03












  • Your answer also works and both gave me some new knowledges. I wish I could mark more than 1 best answers. Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:51
















1














Calling it with



call_user_func_array($name, $params);


it is expecting a standalone function called $name.



As it's a method in a class you need to add this information to the callable, if you want to call it on the current instance then use 



call_user_func_array(array($this,$name), $params);


If it isn't a method in the current instance, then replace $this with the appropriate instance. Or change the method to be static and replace $this with the class name.






share|improve this answer























  • __call and protected function convertPriceByGivenCurrencies are both in CurrencyConverter class. And the variable $currencyConverter is instance of the CurrencyConverter class. Sorry about the unclear explanation :)
    – Toma Tomov
    Nov 13 '18 at 8:03












  • Your answer also works and both gave me some new knowledges. I wish I could mark more than 1 best answers. Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:51














1












1








1






Calling it with



call_user_func_array($name, $params);


it is expecting a standalone function called $name.



As it's a method in a class you need to add this information to the callable, if you want to call it on the current instance then use 



call_user_func_array(array($this,$name), $params);


If it isn't a method in the current instance, then replace $this with the appropriate instance. Or change the method to be static and replace $this with the class name.






share|improve this answer














Calling it with



call_user_func_array($name, $params);


it is expecting a standalone function called $name.



As it's a method in a class you need to add this information to the callable, if you want to call it on the current instance then use 



call_user_func_array(array($this,$name), $params);


If it isn't a method in the current instance, then replace $this with the appropriate instance. Or change the method to be static and replace $this with the class name.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 13 '18 at 7:51

























answered Nov 13 '18 at 7:44









Nigel RenNigel Ren

25.5k61832




25.5k61832












  • __call and protected function convertPriceByGivenCurrencies are both in CurrencyConverter class. And the variable $currencyConverter is instance of the CurrencyConverter class. Sorry about the unclear explanation :)
    – Toma Tomov
    Nov 13 '18 at 8:03












  • Your answer also works and both gave me some new knowledges. I wish I could mark more than 1 best answers. Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:51


















  • __call and protected function convertPriceByGivenCurrencies are both in CurrencyConverter class. And the variable $currencyConverter is instance of the CurrencyConverter class. Sorry about the unclear explanation :)
    – Toma Tomov
    Nov 13 '18 at 8:03












  • Your answer also works and both gave me some new knowledges. I wish I could mark more than 1 best answers. Thank you very much!
    – Toma Tomov
    Nov 13 '18 at 8:51
















__call and protected function convertPriceByGivenCurrencies are both in CurrencyConverter class. And the variable $currencyConverter is instance of the CurrencyConverter class. Sorry about the unclear explanation :)
– Toma Tomov
Nov 13 '18 at 8:03






__call and protected function convertPriceByGivenCurrencies are both in CurrencyConverter class. And the variable $currencyConverter is instance of the CurrencyConverter class. Sorry about the unclear explanation :)
– Toma Tomov
Nov 13 '18 at 8:03














Your answer also works and both gave me some new knowledges. I wish I could mark more than 1 best answers. Thank you very much!
– Toma Tomov
Nov 13 '18 at 8:51




Your answer also works and both gave me some new knowledges. I wish I could mark more than 1 best answers. Thank you very much!
– Toma Tomov
Nov 13 '18 at 8:51


















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