Variable Variables not working right?

[Chris]

If I'm thinking right, this should work...can someone help me
understand why it's not.

//$loop = 1
$test = "num$loop"; //test should now equal "num1"
$ttest = $$test; //ttest should now reference $num1
echo $test; //this echo DOES show "num1"
echo ${$ttest}; //this echo shows nothing (blank)
echo $num1; //this echo shows the contents of $num1
correctly.

What I'm doing is this: From a previous form I have outputted a bunch
of variables: $num1, $num2, $num3...$name1, $name2, $name3.

In my script I want to put in a loop a dynamic reference to each of
these...the step the loop is in will = the last digit in the variable.

What am I missing here? Can someone help?

Thanks,
Chris

[Jay Moore]

Chris wrote:

If I'm thinking right, this should work...can someone help me
understand why it's not.

//$loop = 1
$test = "num$loop"; //test should now equal "num1"
$ttest = $$test; //ttest should now reference $num1
echo $test; //this echo DOES show "num1"
echo ${$ttest}; //this echo shows nothing (blank)
echo $num1; //this echo shows the contents of $num1
correctly.

What I'm doing is this: From a previous form I have outputted a bunch
of variables: $num1, $num2, $num3...$name1, $name2, $name3.

In my script I want to put in a loop a dynamic reference to each of
these...the step the loop is in will = the last digit in the variable.

What am I missing here? Can someone help?

Thanks,
Chris

$num1 = 'hello';
$loop = 1;
$test = "num$loop"; //test should now equal "num1"
$ttest = $$test; //ttest should now reference $num1
echo $test; //this echo DOES show "num1"
echo $ttest; //this echo shows "hello"
echo $num1"; //this echo shows the contents of $num1

This works as intended. Your error was in this line:
echo ${$ttest}; //this echo shows nothing (blank)

$ttest is already referencing another variable. You don't need to
reference it again. In essence, you were echoing $$num1, which, as you
know, is a reference to whatever $num1 is referencing, which is nothing.
In this example, $ttest = $num1 and not $$num1 (which is what you
were trying to echo).

Hope that was clear.

-Jay

[Chris]

> This works as intended. Your error was in this line:

> echo ${$ttest}; //this echo shows nothing (blank)

$ttest is already referencing another variable. You don't need to
reference it again. In essence, you were echoing $$num1, which, as you
know, is a reference to whatever $num1 is referencing, which is nothing.
In this example, $ttest = $num1 and not $$num1 (which is what you
were trying to echo).

Hope that was clear.

-Jay

AHH....thanks Jay. I see what you are saying and it does make sense! :-)

Thanks,
Chris