seperate debug variable in scripts included
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
asked Nov 23 '18 at 7:26
ccalbow
208
add a comment |
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
asked Nov 23 '18 at 7:26
ccalbow
208
Have you tried to debug what$filevalue is? If the point is to produce a different variable name for each script, you shouldn't use another variable$debugValueto 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
add a comment |
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
asked Nov 23 '18 at 7:26
ccalbow
208
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
asked Nov 23 '18 at 7:26
ccalbow
208
asked Nov 23 '18 at 7:26
ccalbow
208
asked Nov 23 '18 at 7:26
ccalbow
208
asked Nov 23 '18 at 7:26
ccalbow
208
asked Nov 23 '18 at 7:26
ccalbow
208
208
Have you tried to debug what$filevalue is? If the point is to produce a different variable name for each script, you shouldn't use another variable$debugValueto 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
add a comment |
Have you tried to debug what$filevalue is? If the point is to produce a different variable name for each script, you shouldn't use another variable$debugValueto 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
add a comment |
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Have you tried to debug what
$filevalue is? If the point is to produce a different variable name for each script, you shouldn't use another variable$debugValueto 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