Variable declaration

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February 13th, 2013, 10:18 PM

gayathri.G

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PHP5 - Variable declaration

Is it possibile to store the php file name in a variable like $a=a.php instead of calling it explicitly like include(a.php). If means how to call it.Any ideas please tell me.

February 13th, 2013, 10:24 PM

Jacques1

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Hi,

not sure what you mean. Yes, you can store the file name in a variable and use that variable like the original name -- if that's what you mean.

PHP Code:


		
			
 $functions_script = 'functions.php';

require_once $functions_script;

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February 13th, 2013, 10:29 PM

gayathri.G

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Thanks Jacques1. You have correctly understood my question. This is the exact one what i'm trying for past few days. Thanks again.

February 13th, 2013, 10:47 PM

gayathri.G

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Please tell me how to declare it in if condition.

PHP Code:


		
			
for($i=0; $i<6;$i++){

if($var[$i]==$str)

{

    $var1="a.php";

    require_once $var1;  

}

else if($var[$i]==$str){

                               $var1="b.php";

                           require_once $var1; 

                               }

}

It is showing some error while doing like this.

February 13th, 2013, 10:53 PM

Jacques1

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What is "some error"? What's the exact error message?

Apart from that, the code is kind of strange. The condition for "if" and "elseif" are exactly the same, which means the "elseif" will never be executed. And requiring constant files in a loop also makes little sense.

What are you trying to do?

February 13th, 2013, 10:59 PM

gayathri.G

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My error message is Fatal error: Cannot redeclare qurexecute() (previously declared in /opt/lampp/htdocs/admin/a.php:1033) in /opt/lampp/htdocs/admin/b.php on line 1031
My table having two columns. I stored one column values in a array. Now i'm checking whether what i'm clicking is in the array or not. If it is seen in array it should open the corresponding php file.Like this i'm trying to do.

PHP5 - Variable declaration

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