PHP: append to value if the key already exist, if not add the key

Question

I am looking for a succinct way of doing this in PHP:

given an array, if I add one key=>value pair to it, the routine should check whether the key already exist.

If it doesn't exist, add to the array with the key=>value pair.

If it does, then the value should be append to the value of the array. So, for example, if the initial array is this

arr['a']='2e'

When I add a 'a'=>'45' pair to the array, then the routine will return me

arr['a']=array('2e', '45')

When I add another 'a=>gt' pair to it, then the routine will return me

arr['a']=array('2e', '45','gt')

Is there a succinct way of doing this? Of course I can write it myself but I believe my solution is very ugly.

You should clarify something, do you really want arr['a']='2e' at the begining, or is it a typo and you expect arr['a']=array('2e') ? — Bite code
– Bite code, Commented Aug 28, 2009 at 10:17

FlorianH · Accepted Answer · 2009-08-28 10:18:29Z

30

You could solve the problem, by using an array for the first element ("2e") aswell:

$arr = array();

$arr['a'][] = '2e';
$arr['a'][] = '45';
$arr['a'][] = 'gt';

print_r($arr);

answered Aug 28, 2009 at 10:18

FlorianH

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1 Comment

Deepak Rajpal Over a year ago

Good that it does not throw any notice/warning if key ['a'] does not exist and eventually creates that key.

VoteyDisciple · Accepted Answer · 2017-12-01 12:09:14Z

14

There are three situations:

The key is undefined
The key is defined, but isn't yet set to an array
The key is defined, and the element is an array.

So, in code:

function appendThings(/* map[string,mixed] */ $array, /* string */ $key, /* string */ $value) {
    if (!isset($array[$key]))
        $array[$key] = $value;
    else if (is_array($array[$key]))
        $array[$key][] = $value;
    else
        $array[$key] = array($array[$key], $value);

    return $array;
}

It's only the last case that's tricky: if it's not an array yet, you'll need to compose one using the current value plus the new one.

edited Dec 1, 2017 at 12:09

answered Aug 28, 2009 at 10:17

VoteyDisciple

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Comments

Amber · Accepted Answer · 2009-08-28 10:19:28Z

3

function update_keypair($arr, $key, $val)
{
   if(empty($arr[$key])) $arr[$key] = array($val);
   else $arr[$key][] = $val;
}

does exactly what you want.

answered Aug 28, 2009 at 10:19

Amber

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RaYell · Accepted Answer · 2009-08-28 10:16:37Z

0

You need to write a function that does that. Or initialize your first element as an array as well and use array_push function to add new elements.

$a = array('2e');
array_push($a, '45');
array_push($a, 'gt');

answered Aug 28, 2009 at 10:16

RaYell

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Comments

mck89 · Accepted Answer · 2009-08-28 10:19:01Z

0

Try this

$key="a";
$value="b";
$array=array();

if(!array_key_exists($key,$array)) $array[$key]=$value;
elseif(is_array($array[$key]))$array[$key][]=$value;
else $array[$key]=array($array[$key],$value);

answered Aug 28, 2009 at 10:19

mck89

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Comments

n1313 · Accepted Answer · 2009-08-28 10:19:51Z

0

if (isset($array[$key]) {
  if (!is_array($array[$key]))
    $array[$key] = (array)$array[$key];
  $array[$key][] = $new_value;
} else {
  $array[$key] = $new_value;
}

Something like that? You can surely simplify this by adding first value as an one-element array, or by using ternar operators, but anyway you'll need a custom function to do the job.

answered Aug 28, 2009 at 10:19

n1313

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Community · Accepted Answer · 2023-11-21 15:10:29Z

0

Strictly array:

$arr['a']=(is_array($arr['a'])? '2e' : array_merge(Array('2e'),$arr['a']));

String with separators:

$arr['a'].='2e'.'/'; // '/' is used as a separator in here.

if you need the string as an array just do $arr['a'] = explode("/",$arr['a']);

both methods are ugly... you should try, as FlorianH suggested, to use the whole variable as an array.

Another method might be to use the Interface in PHp and build something that make suse of the Iterator and ArrayAccess interfaces. (https://www.php.net/manual/en/class.iterator.php, https://www.php.net/manual/en/class.arrayaccess.php)

edited Nov 21, 2023 at 15:10

CommunityBot

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answered Aug 28, 2009 at 11:14

Quamis

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