Graal Forums - makevar with array?

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- - makevar with array? (https://forums.graalonline.com/forums/showthread.php?t=74903)

JkWhoSaysNi

06-26-2007 06:03 PM

makevar with array?

This won't work:

PHP Code:


		
			
makevar("this.something[0]") = "just testing";

echo(this.something[0]);

I could do:

PHP Code:


		
			
makevar("this.something")[0]

but what I really want to do is this:

PHP Code:


		
			
makevar("this.something[0].whatever") = "just testing";

Is there any way to get makevar to work with arrays like this?

Thanks.

Novo	06-26-2007 06:04 PM

Quote:

Originally Posted by JkWhoSaysNi (Post 1322704)

This won't work:

PHP Code:


		
			
makevar("this.something[0]") = "just testing";

echo(this.something[0]);

I could do:

PHP Code:


		
			
makevar("this.something")[0]

but what I really want to do is this:

PHP Code:


		
			
makevar("this.something[0].whatever") = "just testing";

Is there any way to get makevar to work with arrays like this?

Thanks.

Why not use (@"this.something")[0].(@"whatever"); ?

JkWhoSaysNi

06-26-2007 06:15 PM

hmm actually that doesnt seem to work either

PHP Code:


		
			
temp.test = "something";
this.(@temp.test)[0] = 5;
echo(this.something[0]);

echos null.

JkWhoSaysNi

06-26-2007 07:42 PM

Well I see the problem.

"something[0]" becomes a variable name listed in getVarNames() it doesn't treat it as an array at all :(

Novo	06-26-2007 08:11 PM

Quote:

Originally Posted by JkWhoSaysNi (Post 1322711)

hmm actually that doesnt seem to work either

PHP Code:


		
			
temp.test = "something";

this.(@temp.test).add(5);

echo(this.something[0]);

echos null.

PHP Code:


		
			
temp.test = "something";

this.(@temp.test) = new[1];

this.(@temp.test)[0] = 5;

echo(this.something[0]);

I'd try that. As a blank string, I don't think [0] is even created... The variable type isn't set yet. By doing the

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