PHP Dynamically accessing an associative array

Question

EDITED FROM ORIGINAL THAT HAD IMPLIED ONLY 1 ACCESS

If I have an array that contains x number of arrays, each of the form

array('country' => array('city' => array('postcode' => value)))

and another array that might be

array('country', 'city', 'postcode') or array('country', 'city')

depending on what I need to retrieve, how do I use the second array to identify the index levels into the first array and then access it.

I would almost call that a: stealth edit! Please mark it as an EDIT if you extend your question! And if the edit is a completely new question make a new on! — Rizier123
– Rizier123, Commented Dec 17, 2014 at 18:57
It's not a completely new question. It was edited to show that it has to be done multiple times. — Jeff Greenberg
– Jeff Greenberg, Commented Dec 17, 2014 at 18:58
sorry but this edit is the same of your original question, you can't have more than 1 value inside $array['capital']['city']['postcode'] as I have explained in comments — dynamic
– dynamic, Commented Dec 17, 2014 at 18:59
No, you can't. But that wasn't the point. I was simply saying that if the array contains 20 countries, do I need to iterate again and again based on the indexes for each of the 20, or is there a way to assemble the indexes into a variable variable and access each of the 20 using it and not iterating. — Jeff Greenberg
– Jeff Greenberg, Commented Dec 17, 2014 at 19:01

dynamic · Accepted Answer · 2014-12-17 18:44:36Z

1

By nesting references with $cur = &$cur[$v]; you can read and modify the original value:
Live example on ide1: http://ideone.com/xtmrr8

$array = array('x' => array('y' => array('z' => 20)));
$keys = array('x', 'y', 'z');

// Start nesting new keys
$cur = &$array;
foreach($keys as $v){
    $cur = &$cur[$v];
}

echo $cur;  // prints 20
$cur = 30;  // modify $array['x']['y']['z']

edited Dec 17, 2014 at 18:44

answered Dec 17, 2014 at 18:37

dynamic

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6 Comments

Rizier123 Over a year ago

I think that's the best possible answer! Very nice explained and good comments! Deserves an upvotes :D

Jeff Greenberg Over a year ago

Let me add one more consideration. If I have 20 of $array, say each being a country, and given $keys, I need to access each of the 20 in the same way, is there a way to do it without iterating each?

dynamic Over a year ago

I don't know if that is a different question or is it related. Anyway if this answer helped you consider to +1 it

Jeff Greenberg Over a year ago

Same question. Just trying to consider efficiency. If my keys are ('capital','city','postcode']) and I want the postcode in each of 20 countries, once I determine the keys, is iterating as such what I should do? Could be, that there is no way to do it as a variable variable (like ${$myarray}{['x']['y']['z']} or something to that effect. Just making sure.

dynamic Over a year ago

how can you have 20 countries if you are accesing an exact item in your array $array['capital']['city']['postcode'] ? Anyway I think this is another question open a new question

|

Glen · Accepted Answer · 2014-12-17 18:46:23Z

1

Loop through the array of indexes, traveling down the initial array step by step until you reach the end of the index array.

$array1 = array('x' => array('y' => array('z' => 20)));
$keys = array('x', 'y', 'z');
$array_data = &$array1;
foreach($keys as $key){
    $array_data = &$array_data[$key];
}
echo $array_data;

edited Dec 17, 2014 at 18:46

answered Dec 17, 2014 at 18:39

Glen

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2 Comments

dynamic Over a year ago

this will create a new copy of that value

Glen Over a year ago

Edited to pass by reference

Jeff Greenberg · Accepted Answer · 2014-12-17 20:43:26Z

0

Ok, I apologize for not having expressed my question more clearly, originally, but I got it to work as I was hoping, with a variable variable, as follows:

$keystring = '';
foreach ($keys as $key) {
  $keystring .= "['$key']";
}

Then iterate the main array, for each of the country entries in it, and access the desired value as:

eval('$value = $entry' . "$keystring;");

edited Dec 17, 2014 at 20:43

answered Dec 17, 2014 at 19:31

Jeff Greenberg

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