seperate debug variable in scripts included












0














Additional to my last question I have a new problem:
I would like to have the same debug code on each script but cannot avoid the included page changing the value of the debug variable: See this:
php file "loadData.php" hast the following debug script:



$file=pathinfo(__FILE__)['filename'];

$debugValue=true;

$debug{$file}=$debugValue;



if($debug{$file}){

    echo "Achtung: Testausgabe von ".$file.".php ist an!</br>"; 

}



$ini_config = parse_ini_file("config.ini", true);

$DB=$ini_config['php']['database'];//Speicherung der Daten



//database einbinden

include($DB);



//FUNKTION getProjektID:





if($debug{$file}){

    $projektname='Testprojekt';

}else{

    $projektname=$_REQUEST['projektname'];

}


I am including the file database.php with include($DB);.



Within the database.php I have the following debug code:



$file=pathinfo(__FILE__)['filename'];

$debugValue=false;



$debug{$file}=$debugValue;



if($debug{$file}){

    echo "Achtung: Testausgabe von ".$file.".php ist an!</br>"; 

}


My problem now is that the $debug{$file}-Value changes to false after the inculsion of the database.php...



I thought with my workaround creating a dynamic variable in each file, i can avoid that the debug variables are being changed by another script, but this does not seem to work.



How can I avoid this?



Is there a way to fix the variable $debug{$file} within a script so that it cannot be changed after including another script???



Thanks for helping.






php include dynamic-variables







share|improve this question






















  • Have you tried to debug what $file value is? If the point is to produce a different variable name for each script, you shouldn't use another variable $debugValue to set the value, as it makes the thing pointless by overriding the same variable for each script. Just set $debug{$file} directly
    – Kaddath
    Nov 23 '18 at 7:41


















0














Additional to my last question I have a new problem:
I would like to have the same debug code on each script but cannot avoid the included page changing the value of the debug variable: See this:
php file "loadData.php" hast the following debug script:



$file=pathinfo(__FILE__)['filename'];

$debugValue=true;

$debug{$file}=$debugValue;



if($debug{$file}){

    echo "Achtung: Testausgabe von ".$file.".php ist an!</br>"; 

}



$ini_config = parse_ini_file("config.ini", true);

$DB=$ini_config['php']['database'];//Speicherung der Daten



//database einbinden

include($DB);



//FUNKTION getProjektID:





if($debug{$file}){

    $projektname='Testprojekt';

}else{

    $projektname=$_REQUEST['projektname'];

}


I am including the file database.php with include($DB);.



Within the database.php I have the following debug code:



$file=pathinfo(__FILE__)['filename'];

$debugValue=false;



$debug{$file}=$debugValue;



if($debug{$file}){

    echo "Achtung: Testausgabe von ".$file.".php ist an!</br>"; 

}


My problem now is that the $debug{$file}-Value changes to false after the inculsion of the database.php...



I thought with my workaround creating a dynamic variable in each file, i can avoid that the debug variables are being changed by another script, but this does not seem to work.



How can I avoid this?



Is there a way to fix the variable $debug{$file} within a script so that it cannot be changed after including another script???



Thanks for helping.






php include dynamic-variables







share|improve this question






















  • Have you tried to debug what $file value is? If the point is to produce a different variable name for each script, you shouldn't use another variable $debugValue to set the value, as it makes the thing pointless by overriding the same variable for each script. Just set $debug{$file} directly
    – Kaddath
    Nov 23 '18 at 7:41
















0












0








0







Additional to my last question I have a new problem:
I would like to have the same debug code on each script but cannot avoid the included page changing the value of the debug variable: See this:
php file "loadData.php" hast the following debug script:



$file=pathinfo(__FILE__)['filename'];

$debugValue=true;

$debug{$file}=$debugValue;



if($debug{$file}){

    echo "Achtung: Testausgabe von ".$file.".php ist an!</br>"; 

}



$ini_config = parse_ini_file("config.ini", true);

$DB=$ini_config['php']['database'];//Speicherung der Daten



//database einbinden

include($DB);



//FUNKTION getProjektID:





if($debug{$file}){

    $projektname='Testprojekt';

}else{

    $projektname=$_REQUEST['projektname'];

}


I am including the file database.php with include($DB);.



Within the database.php I have the following debug code:



$file=pathinfo(__FILE__)['filename'];

$debugValue=false;



$debug{$file}=$debugValue;



if($debug{$file}){

    echo "Achtung: Testausgabe von ".$file.".php ist an!</br>"; 

}


My problem now is that the $debug{$file}-Value changes to false after the inculsion of the database.php...



I thought with my workaround creating a dynamic variable in each file, i can avoid that the debug variables are being changed by another script, but this does not seem to work.



How can I avoid this?



Is there a way to fix the variable $debug{$file} within a script so that it cannot be changed after including another script???



Thanks for helping.






php include dynamic-variables







share|improve this question













Additional to my last question I have a new problem:
I would like to have the same debug code on each script but cannot avoid the included page changing the value of the debug variable: See this:
php file "loadData.php" hast the following debug script:



$file=pathinfo(__FILE__)['filename'];

$debugValue=true;

$debug{$file}=$debugValue;



if($debug{$file}){

    echo "Achtung: Testausgabe von ".$file.".php ist an!</br>"; 

}



$ini_config = parse_ini_file("config.ini", true);

$DB=$ini_config['php']['database'];//Speicherung der Daten



//database einbinden

include($DB);



//FUNKTION getProjektID:





if($debug{$file}){

    $projektname='Testprojekt';

}else{

    $projektname=$_REQUEST['projektname'];

}


I am including the file database.php with include($DB);.



Within the database.php I have the following debug code:



$file=pathinfo(__FILE__)['filename'];

$debugValue=false;



$debug{$file}=$debugValue;



if($debug{$file}){

    echo "Achtung: Testausgabe von ".$file.".php ist an!</br>"; 

}


My problem now is that the $debug{$file}-Value changes to false after the inculsion of the database.php...



I thought with my workaround creating a dynamic variable in each file, i can avoid that the debug variables are being changed by another script, but this does not seem to work.



How can I avoid this?



Is there a way to fix the variable $debug{$file} within a script so that it cannot be changed after including another script???



Thanks for helping.






php include dynamic-variables




php include dynamic-variables






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share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '18 at 7:26









ccalbow

208




208












  • Have you tried to debug what $file value is? If the point is to produce a different variable name for each script, you shouldn't use another variable $debugValue to set the value, as it makes the thing pointless by overriding the same variable for each script. Just set $debug{$file} directly
    – Kaddath
    Nov 23 '18 at 7:41




















  • Have you tried to debug what $file value is? If the point is to produce a different variable name for each script, you shouldn't use another variable $debugValue to set the value, as it makes the thing pointless by overriding the same variable for each script. Just set $debug{$file} directly
    – Kaddath
    Nov 23 '18 at 7:41


















Have you tried to debug what $file value is? If the point is to produce a different variable name for each script, you shouldn't use another variable $debugValue to set the value, as it makes the thing pointless by overriding the same variable for each script. Just set $debug{$file} directly
– Kaddath
Nov 23 '18 at 7:41






Have you tried to debug what $file value is? If the point is to produce a different variable name for each script, you shouldn't use another variable $debugValue to set the value, as it makes the thing pointless by overriding the same variable for each script. Just set $debug{$file} directly
– Kaddath
Nov 23 '18 at 7:41














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