php (complex) dynamic variable name construction

Question

I've got this problem when trying to construct a variable name (which should output a corresponding array element.)

I start with setting the variables for testing purposes, normally the piece variable would hold the first element of an array $piece.

$wordnumber = 0;
$piece[0] = "forever";

I later echo them right before my problem to see if they're still ok.

echo "$piece[0]";
echo "$wordnumber";

The output is ok.

forever0

But then comes the problem, as I'm trying to make a function that automatically handles every single array element, so I want it to construct the next corresponding variable every time. However somehow it has no value after construction.

$name = ${'piece[' . $wordnumber . ']'};
echo "$name";

outputs nothing...

I've tried a lot of different formatting, I really don't know why I'm failing so hard here. The code isn't part of any function right now btw.

Update: $name = $piece[$wordnumber] solves the problem I'm curious though why my previous formatting didn't work as expected.

Update: Question solved by André, the problem was that $piece[0] wasn't actually part of an array. So $piece was the actual variable. After storing an actual array $piece = array("Redish", "Yellow", "Green"); at start and using global $piece; in my function everything started working like a charm.

possible duplicate. PHPs variable variables syntax doesn't support access by array index. read my full my answer (with solution) here: stackoverflow.com/questions/9374292/variable-name-as-variable/… — Kaii
– Kaii, Commented Feb 27, 2012 at 18:45
It seems I didn't search existing questions well enough just found this formatting $myTempArr[ $key ]; on link So I changed it to $name = $piece[ $wordnumber ]; and now it works like a charm outputting forever. — Happy
– Happy, Commented Feb 27, 2012 at 18:56
My question remains though, why ${piece[0]} with my apparently somehow faulty construction didn't work. I'm the curious type I guess. — Happy
– Happy, Commented Feb 27, 2012 at 19:01

André · Accepted Answer · 2012-02-27 19:41:16Z

1

I haven’t tested the code, but I think your first approach didn’t work because your variable’s name is “piece”, not “piece[0]”. In other words, ${'piece'}[0] should work, but ${'piece[0]'} is wrong. Try add this in the very beginning of your script and PHP should display you some complaints:

error_reporting(E_ALL);
ini_set("display_errors", 1);

answered Feb 27, 2012 at 19:41

André

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2 Comments

Happy Over a year ago

Very good André! Smart thinking! And very educational for me. The problem was indeed in the variable not being an actual array! I changed it to $piece = array("Red", "Yellow", "Green", "Blue", "Indigo"); and now everything works like a charm.

André Over a year ago

Thanks for your kind words. If my answer was enough for you, please set it as your accepted answer ;]

Samuel Herzog · Accepted Answer · 2012-02-27 19:00:32Z

1

Try just $name = $piece[$wordnumber]; and echo $name
this will output "forever"

answered Feb 27, 2012 at 19:00

Samuel Herzog

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2 Comments

Happy Over a year ago

Yes this is correct, it also works with spaces like this $piece[ $wordnumber ] btw. I'm meanwhile searching for the reason behind my first formatting's dysfunction.

Samuel Herzog Over a year ago

the array dereferers ([]) shall not be included in a string I think. Therefore ${'parts'}[$wordnumber] works.

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