galexcd
Oct 25 2007, 08:31 PM
| | 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:
CODE $test=array(1,2,5);
$name="test";
//Doesn't work:
echo $$name[0];
/Doesn't work:
echo ${$name}[0]; |
Reply
jlhaslip
Oct 25 2007, 10:38 PM
CODE <?php $test=array(1,2,5); $name = &$test; // check this line echo 'print_r $test'; echo '<br />'; echo '<pre>'; print_r ($test); echo '</pre>'; echo '<br />'; echo 'print_r $test[0]'; echo '<br />'; print_r ($test[0]); echo '<br />'; echo '<br />'; echo 'print_r $name'; echo '<pre>'; print_r ($name); echo '</pre>'; echo '<br />'; echo 'print_r $name[0]'; echo '<pre>'; print_r ($name[0]); echo '</pre>'; ?> Note line 2 and the prefix for references, compared to your code.
Reply
galexcd
Oct 26 2007, 01:32 AM
I was hoping not to use references but all right i guess i must...
Reply
pointybirds
Oct 26 2007, 02:22 PM
Alex, I believe the evil eval could be your friend here: CODE $test=array(1,2,5);
$name="test";
eval ("\$blah = $${name}[2];");
echo $blah; Use it with care and good luck!
Reply
galexcd
Oct 26 2007, 05:12 PM
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.
Reply
pointybirds
Oct 26 2007, 11:12 PM
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.
Reply
i test you code, work. my version php 5.2 CODE <?php
$test=array(1,2,5);
$name="test";
echo ${$name}[2]; // print 5 ?!
?>
Reply
galexcd
Nov 5 2007, 05:46 PM
Hmm it didn't work for me... Perhaps its because I have an older version?
Reply
jlhaslip
Nov 5 2007, 06:00 PM
works for me, too php 5.2.4
Reply
you can see php version if you run <?php phpinfo(); ?>
Reply
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