call_user_func_array throws an error when the method exist
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
asked Nov 13 '18 at 7:38
Toma TomovToma Tomov
612216
add a comment |
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
asked Nov 13 '18 at 7:38
Toma TomovToma Tomov
612216
When you are checking whether the method name exists, you are passing a class name to do so. But where in yourcall_user_func_arraycall 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
add a comment |
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
asked Nov 13 '18 at 7:38
Toma TomovToma Tomov
612216
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
asked Nov 13 '18 at 7:38
Toma TomovToma Tomov
612216
asked Nov 13 '18 at 7:38
Toma TomovToma Tomov
612216
asked Nov 13 '18 at 7:38
Toma TomovToma Tomov
612216
asked Nov 13 '18 at 7:38
Toma TomovToma Tomov
612216
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 yourcall_user_func_arraycall 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
add a comment |
When you are checking whether the method name exists, you are passing a class name to do so. But where in yourcall_user_func_arraycall 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
add a comment |
2 Answers
2
active
oldest
votes
$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.
So to call it, assuming it is a static method, you would need something like:
CurrencyConverter::$name(...$params);
Note that you need the ... operator to unpack $params
answered Nov 13 '18 at 7:44
jeroenjeroen
81.8k1698122
But the unpacking is avaible in 5.6 only. At least this is what the PHPStorm says :)
– Toma Tomov
Nov 13 '18 at 8:04
@TomaTomov That's right, if you are on an earlier version you should probably upgrade :-)
– jeroen
Nov 13 '18 at 8:09
In fact I am onPHP Version 7.2.6-1. it is not usable on higher version ?
– Toma Tomov
Nov 13 '18 at 8:12
@TomaTomov It is available from 5.6 on, so you can use it without any problems. Does PHPStorm know you are on 7.2?
– jeroen
Nov 13 '18 at 8:28
1
Guess he doesn't know because it is working :) Thank you very much!
– Toma Tomov
Nov 13 '18 at 8:41
add a comment |
Calling it with
call_user_func_array($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 $this with the class name.
answered Nov 13 '18 at 7:44
Nigel RenNigel Ren
25.5k61832
__callandprotected function convertPriceByGivenCurrenciesare both inCurrencyConverterclass. And the variable$currencyConverteris instance of theCurrencyConverterclass. 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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.
So to call it, assuming it is a static method, you would need something like:
CurrencyConverter::$name(...$params);
Note that you need the ... operator to unpack $params
answered Nov 13 '18 at 7:44
jeroenjeroen
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 onPHP 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
add a comment |
$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.
So to call it, assuming it is a static method, you would need something like:
CurrencyConverter::$name(...$params);
Note that you need the ... operator to unpack $params
answered Nov 13 '18 at 7:44
jeroenjeroen
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 onPHP 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
add a comment |
$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.
So to call it, assuming it is a static method, you would need something like:
CurrencyConverter::$name(...$params);
Note that you need the ... operator to unpack $params
answered Nov 13 '18 at 7:44
jeroenjeroen
81.8k1698122
$name by itself is not a known function; it seems to be a method in the CurrencyConverter class.
So to call it, assuming it is a static method, you would need something like:
CurrencyConverter::$name(...$params);
Note that you need the ... operator to unpack $params
answered Nov 13 '18 at 7:44
jeroenjeroen
81.8k1698122
edited Nov 13 '18 at 7:49
answered Nov 13 '18 at 7:44
jeroenjeroen
81.8k1698122
answered Nov 13 '18 at 7:44
jeroenjeroen
81.8k1698122
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 onPHP 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
add a comment |
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 onPHP 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
add a comment |
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 $this with the class name.
answered Nov 13 '18 at 7:44
Nigel RenNigel Ren
25.5k61832
__callandprotected function convertPriceByGivenCurrenciesare both inCurrencyConverterclass. And the variable$currencyConverteris instance of theCurrencyConverterclass. 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
add a comment |
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 $this with the class name.
answered Nov 13 '18 at 7:44
Nigel RenNigel Ren
25.5k61832
__callandprotected function convertPriceByGivenCurrenciesare both inCurrencyConverterclass. And the variable$currencyConverteris instance of theCurrencyConverterclass. 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
add a comment |
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 $this with the class name.
answered Nov 13 '18 at 7:44
Nigel RenNigel Ren
25.5k61832
Calling it with
call_user_func_array($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 $this with the class name.
answered Nov 13 '18 at 7:44
Nigel RenNigel Ren
25.5k61832
edited Nov 13 '18 at 7:51
answered Nov 13 '18 at 7:44
Nigel RenNigel Ren
25.5k61832
answered Nov 13 '18 at 7:44
Nigel RenNigel Ren
25.5k61832
answered Nov 13 '18 at 7:44
Nigel RenNigel Ren
25.5k61832
25.5k61832
__callandprotected function convertPriceByGivenCurrenciesare both inCurrencyConverterclass. And the variable$currencyConverteris instance of theCurrencyConverterclass. 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
add a comment |
__callandprotected function convertPriceByGivenCurrenciesare both inCurrencyConverterclass. And the variable$currencyConverteris instance of theCurrencyConverterclass. 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
add a comment |
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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_arraycall 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