Getting An Array Value Of A Dynamic Variable

[galexcd 0]

Ok i can't believe I need help in php but I can't figure this out... If i have an array, lets say called test, how do i get a certain value from that array when referencing it dynamically by either a string or another variable?

Example:

$test=array(1,2,5);$name="test";//Doesn't work:echo $$name[0];/Doesn't work:echo ${$name}[0];

[jlhaslip 4]

<?php$test=array(1,2,5);$name = &$test; // check this lineecho 'print_r $test';echo '<br />';echo '<pre>';print_r ($test);echo '</pre>';echo '<br />';echo 'print_r $test[0]';echo '<br />';print_r ($test[0]);echo '<br />';echo '<br />';echo 'print_r $name';echo '<pre>';print_r ($name);echo '</pre>';echo '<br />';echo 'print_r $name[0]';echo '<pre>';print_r ($name[0]);echo '</pre>';?>

Note line 2 and the prefix for references, compared to your code.

[galexcd 0]

I was hoping not to use references but all right i guess i must...

[pointybirds 0]

Alex,

I believe the evil eval could be your friend here:

$test=array(1,2,5);$name="test";eval ("\$blah = $${name}[2];");echo $blah;

Use it with care and good luck!

[galexcd 0]

Does eval parse differently than regular php code? I always assumed it was the same. I shall try this and report back here if it works.

[pointybirds 0]

eval will run the contents of the string as code, however it interpolates variables too, so the above code will replace $name with "test", then uses $test in the eval.

[pop 0]

i test you code, work. my version php 5.2

<?php$test=array(1,2,5);$name="test";echo ${$name}[2]; // print 5 ?!?>

[galexcd 0]

Hmm it didn't work for me... Perhaps its because I have an older version?

[jlhaslip 4]

works for me, toophp 5.2.4

[pop 0]

you can see php version if you run<?php phpinfo(); ?>