Add JSON keys/values while creating another JSON PHP

Question

I am trying to pass some JSON keys/values that I have to another JSON I am creating dynamically.

For example, this is the JSON I have in $json_create

{
"Key 1":"Value 1",
"Key 2":"Value 2",
"Key 3":"Value 3",
"Key 4":"Value 4"
}

That comes over file_get_contents

$requestBody = file_get_contents('php://input');
$json_create= json_decode($requestBody);

And this is the JSON I am creating

$final_json = [
       $json_create,
       "type" => "message",
       "elements" => $json_merge,  
       "actions" => $actions_json
  ];

echo json_encode($final_json, JSON_PRETTY_PRINT | JSON_UNESCAPED_UNICODE | JSON_UNESCAPED_SLASHES | JSON_NUMERIC_CHECK);

Which print something like

{
  "type": "message",
  "elements": [
    {
      "msg_text": "This is a simple response message"
    }
  ]
}

What I am trying to achieve is

{
    "Key 1":"Value 1",
    "Key 2":"Value 2",
    "Key 3":"Value 3",
    "Key 4":"Value 4",
    "type": "message",
    "elements": [
        {
          "msg_text": "This is a simple response message"
     }
   ]
 }

There is quite a lot on that subject but somehow I could not succeed in implementing it.

Use json_decode($requestBody, true); to get an array - you are currently getting an object back. — Nigel Ren
– Nigel Ren, Commented Oct 15, 2020 at 17:18

Barmar · Accepted Answer · 2020-10-15 16:51:45Z

2

Use array_merge rather than putting the $json_create inside the array.

$final_json = array_merge($json_create, 
    [
        "type" => "message",
        "elements" => $json_merge,  
        "actions" => $actions_json
    ]);

answered Oct 15, 2020 at 16:51

Barmar

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10 Comments

lStoilov Over a year ago

I am getting bull if I try to merge.

Barmar Over a year ago

You mean getting null?

lStoilov Over a year ago

Yes, sorry typo... NULL

Barmar Over a year ago

That sounds like json_encode() is getting an error. What does json_last_error_message() return?

Barmar Over a year ago

A common reason for that is a character that can't be encoded properly.

|

Biswajit Biswas · Accepted Answer · 2020-10-15 17:05:39Z

1

<?php

$json_create = '{"Key1": "Value 1", "Key2": "Value 2", "Key3": "Value 3", "Key4": "Value 4"}';
$json_create = (array) json_decode($json_create);

$your_array = [
  "type" => "message",
  "elements" => 'foo',
  "actions" => 'bar'
];

$final_json = array_merge($json_create, $your_array);

$result = json_encode($final_json);


echo $result;

output

{
  Key1: "Value 1",
  Key2: "Value 2",
  Key3: "Value 3",
  Key4: "Value 4",
  type: "message",
  elements: "foo",
  actions: "bar"
}

answered Oct 15, 2020 at 17:05

Biswajit Biswas

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3 Comments

lStoilov Over a year ago

Thanks @Biswajit, that did the trick. Particularly the (array) portion $json_create = (array) json_decode($json_create);

Biswajit Biswas Over a year ago

that's called type casting

Barmar Over a year ago

Give a second argument true to json_decode, then it returns an associative array instead of object.

Al-Amin · Accepted Answer · 2020-10-15 18:12:56Z

1

This can be done using union operator also.

$requestBody = '{
    "Key 1":"Value 1",
    "Key 2":"Value 2",
    "Key 3":"Value 3",
    "Key 4":"Value 4"
    }';
$json_create= json_decode($requestBody, true );
$other_array = [
    "type" => "message",
    "elements" => [],  
    "actions" => []
];
$final_json = $json_create + $other_array;

answered Oct 15, 2020 at 18:12

Al-Amin

6564 silver badges17 bronze badges

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