Merging two JSON objects with PHP?

Question

I have two JSON objects, and I want it to be merged into one JSON object. I don't want the JSON data to be overwritten. I want the data to be merged.

Object #1: (example1.json)

[{
    "title": "xx",        
    "releaseYear": "2014",        
},
{
    "title": "yy",       
    "releaseYear": "2015"        
}]

Object #2: (example2.json)

[{
    "title": "yy",       
    "releaseYear": "2017"

}]

This is the PHP File I currently am using:

    <?php

    $jsonString = file_get_contents('example.json');
    $jsonString2 = file_get_contents('example2.json');   
    $data = json_decode($jsonString, true);
    $data2 = json_decode($jsonString2, true);      


    $op = array_merge_recursive( $data, $data2 );

    $resJson = json_encode($op);
    file_put_contents('example3.json', $resJson);

Output received: (example3.json)

[  
   {  
      "title":"xx",
      "releaseYear":"2014"
   },
   {  
      "title":"yy",
      "releaseYear":"2015"
   },
   {  
      "title":"yy",
      "releaseYear":"2017"
   }
]

Required Output:

[{
    "title": "xx",        
    "releaseYear": "2014"        
},
{
    "title": "yy",       
    "releaseYear": "2017"        
}]

Solutions appreciated. Am new to Stackoverflow, so sorry for my bad formatting, if any. Thanks in advance :)

Release year doesn't match on given objects. Is this correct? If so, does this mean you always have to take higher release year value on matching titles when doing merge? — Evaldas Buinauskas
– Evaldas Buinauskas, Commented Oct 25, 2015 at 14:19
I'm confused, you don't want to overwrite data, you want to merge, but the result you're asking for is clearly overwriting, not just merging ? — adeneo
– adeneo, Commented Oct 25, 2015 at 14:23
Overwritten: [{ "title": "yy", "releaseYear": "2017" }] Since there is only one object, it overwrites, and merges the second data alone. This is what I meant as overwriting. — Anirudh Murali
– Anirudh Murali, Commented Oct 25, 2015 at 14:29

VolkerK · Accepted Answer · 2015-10-25 14:23:38Z

1

Many possible solutions. One example:

<?php
$data = json_decode('[{"title": "xx","releaseYear": "2014"},{"title": "yy","releaseYear": "2015"}]', true);
$data2 = json_decode('[{"title": "yy","releaseYear": "2017"}]', true);

$data = foo($data, 'title');
$data = foo($data2, 'title', $data);
$data = array_values($data);
var_export($data);

function foo($src, $prop, $target=array()) {
    foreach($src as $o) {
        $target[ $o[$prop] ] = $o;
    }
    return $target;
}

answered Oct 25, 2015 at 14:23

VolkerK

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3 Comments

Anirudh Murali Over a year ago

First of all, thanks for your answer. Well, I ran it, and it gave me this output: array ( 0 => array ( 'title' => 'xx', 'releaseYear' => '2014', ), 1 => array ( 'title' => 'yy', 'releaseYear' => '2017', ), ) That's what I need. But can I get it in JSON format? Like: [{ "title": "xx", "releaseYear": "2014" }, { "title": "yy", "releaseYear": "2017" }] Later, this data will be parsed in Android with Volley, it doesn't recognize that type of data, that's why. Thanks in advance :)

Anirudh Murali Over a year ago

Thanks for the solution @VolkerK. A simple google helped me to get it in JSON format. Used echo json_encode($data); instead of var_export($data); Thanks again, if you're free, do tell me the use of using var_export :)

VolkerK Over a year ago

I just like the output better than that of var_dump ;-) You can copy the output of var_export and paste it (as-is) back into a php script to use it as sample data.

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