PHP Only Returning Last JSON Object

Question

I'm attempting to do something I've done many times before (access objects in a JSON file with PHP) and for some reason json_decode is only returning the last item in the JSON array. Here's the JSON:

{
  "person": {
    "lname": "smith",
    "fname": "bob"
  },
  "person": {
    "lname": "jones",
    "fname": "jane"
  }
}

And the PHP:

<?php

//access and dump
$json = file_get_contents('people.json');
$filey = json_decode($json, true);

var_dump($filey);

?>

The result is only the last item in the array:

array (size=1)
  'person' => 
    array (size=2)
      'lname' => string 'jones' (length=5)
      'fname' => string 'jane' (length=4)

Using json_last_error returns no errors and I'm valid according to jsonlint. I'm also not finding any console errors when I load the page.

I'm totally stumped and can't see anything different from the times I've done this before - can anyone identify what I'm missing here?

Your indexes are same and that's why it's over-written (person) — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Jul 27, 2016 at 4:39
You don't even need object names in this case. Take a look at my answer. — BeetleJuice
– BeetleJuice, Commented Jul 27, 2016 at 5:07

jarvo69 · Accepted Answer · 2016-07-27 04:35:43Z

2

That's because your json object names "person" within json array are similar so json decode will override the values with latest. Consider something like

{
  "person1": {
    "lname": "smith",
    "fname": "bob"
  },
  "person2": {
    "lname": "jones",
    "fname": "jane"
  }
}

and your code will work fine.

answered Jul 27, 2016 at 4:35

jarvo69

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1 Comment

Marcatectura Over a year ago

Well that's embarrassing - in the past, I haven't used object names and that's why I never realized they needed to be unique. Thanks for your help.

BeetleJuice · Accepted Answer · 2016-07-27 17:40:01Z

Marcatectura, I know you've already accepted the answer that suggests using different object keys but I thought you should know. If you want an array in PHP, You don't even need object names. The following JSON will do:

[
    {
        "lname": "Dawes",
        "fname": "April"
    },
    {
        "lname": "Colin",
        "fname": "Dick"
    }
]

A trick I use when I'm designing my JSON is to build a sample PHP array in the shape I want json_decode to give me, encode that array and output the result to screen. So what I posted above is the result of:

$arr = [
    ['lname'=>'Dawes','fname'=>'April'],['lname'=>'Colin','fname'=>'Dick'],
];
$json = json_encode($arr);
echo $json;

Since I built a JSON by encoding an array having the shape I want, I can be confident that even as my data change, json_decode($json,true) will give me the array shape I expect.

Happy coding.

Lian · Accepted Answer · 2016-07-27 04:52:21Z

0

When you use json_decode(true), your json is now an array. You cannot have two array keys that are the same, in this case "person".

If you still want to use json_decode(true), then change "person" to "person1" or so.

Try both var_dump($filey) and var_dump($json), you will see what I'm talking about.

answered Jul 27, 2016 at 4:52

Lian

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