json_encode encodes a sparse array as a JSON object

Question

I'm experiencing odd behavior with json_encode after removing a numeric array key with unset:

$a = array(
    new stdclass,
    new stdclass,
    new stdclass
);
$a[0]->abc = '123';
$a[1]->jkl = '234';
$a[2]->nmo = '567';

printf("%s\n", json_encode($a));
unset($a[1]);
printf("%s\n", json_encode($a));

Program Output

[{"abc":"123"},{"jkl":"234"},{"nmo":"567"}]
{"0":{"abc":"123"},"2":{"nmo":"567"}}

The first time $a is converted to JSON, it's encoded as an array. The second time, it's encoded as an object. Why is this happening, and how can I prevent it?

zneak · Accepted Answer · 2010-10-06 02:33:37Z

22

The reason for that is that your array has a hole in it: it has the indices 0 and 2, but misses 1. JSON can't encode arrays with holes because the array syntax has no support for indices.

You can encode array_values($a) instead, which will return a reindexed array.

answered Oct 6, 2010 at 2:33

zneak

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4 Comments

NullUserException Over a year ago

Alternatively you can sort the array after unsetting an element.

zneak Over a year ago

@NullUserException Though that will have the side effect of, well, sorting your array.

leepowers Over a year ago

For those wondering about the unset behavior, view the following link and scroll down to "Useful functions": us3.php.net/manual/en/language.types.array.php "The unset() function allows removing keys from an array. Be aware that the array will not be reindexed"

sjkm Over a year ago

@NullUserException Great addition! Needed to sort the array anyway - killing two birds with one stone.

leepowers · Accepted Answer · 2010-10-06 02:55:52Z

7

In addition to the array_values technique it is possible to use array_splice and remove an element and re-index in one step:

unset($a[1]);

Instead:

array_splice($a, 1, 1);

answered Oct 6, 2010 at 2:55

leepowers

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2 Comments

CookieMonster Over a year ago

absolutely perfect way to do the job without messing up the collection, thanks!

DQM Over a year ago

This. The most elegant way.

barbushin · Accepted Answer · 2010-10-08 11:18:24Z

2

Try to use JSON_FORCE_OBJECT option for json_encode, like: json_encode($a, JSON_FORCE_OBJECT) so you will always have the same result.

answered Oct 8, 2010 at 11:18

barbushin

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