How to extract and access data from JSON with PHP?

Solution:

Intro

First off you have a string. JSON is not an array, an object, or a data structure. JSON is a text-based serialization format - so a fancy string, but still just a string. Decode it in PHP by using .

 {-code-2}
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• scalars: strings, ints, floats, and bools
• {-code-25}s (a special type of its own)
• compound types: objects and arrays.

These are the things that can be encoded in JSON. Or more accurately, these are PHP's versions of the things that can be encoded in JSON.

There's nothing special about them. They are not "JSON objects" or "JSON arrays." You've decoded the JSON - you now have basic everyday PHP types.

Objects will be instances of stdClass, a built-in class which is just a generic thing that's not important here.

Accessing object properties

You access the properties of one of these objects the same way you would for the public non-static properties of any other object, e.g.{-code-3}.

{-code-4}
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Accessing array elements

You access the elements of one of these arrays the same way you would for any other array, e.g. .

{-code-6}
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Iterate over it with .

{-code-7} ($toppings as $topping) {
    echo $topping, "\n";
}
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Or mess about with any of the bazillion built-in array functions.

Accessing nested items

The properties of objects and the elements of arrays might be more objects and/or arrays - you can simply continue to access their properties and members as usual, e.g.{-code-9}.

{-code-10}
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Passing{-code-11} as the second argument to {-code-1}

When you do this, instead of objects you'll get associative arrays - arrays with strings for keys. Again you access the elements thereof as usual, e.g.{-code-12}.

$json = '
{
    "type": "donut",
    "name": "Cake",
    "toppings": [
        { "id": "5002", "type": "Glazed" },
        { "id": "5006", "type": "Chocolate with Sprinkles" },
        { "id": "5004", "type": "Maple" }
    ]
}';

$yummy = json_decode($json, {-code-11});

echo $yummy['toppings'][2]['type']; //Maple
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Accessing associative array items

When decoding a JSON object to an associative PHP array, you can iterate both keys and values using the syntax, eg

$json = '
{
    "foo": "foo value",
    "bar": "bar value",
    "baz": "baz value"
}';

$assoc = json_decode($json, {-code-11});
{-code-7} ($assoc as $key => $value) {
    echo "The value of key '$key' is '$value'", PHP_EOL;
}
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Hit the decoded data with a :

{-code-19}
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{-code-20}
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{-code-22}
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{-code-23}
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Take a look at it in this handy interactive JSON explorer.

Break the problem down into pieces that are easier to wrap your head around.

{-code-1} returns{-code-25}

This happens because either:

1. The JSON consists entirely of just that,{-code-25}.
2. The JSON is invalid - check the result of or put it through something like JSONLint.
3. It contains elements nested more than 512 levels deep. This default max depth can be overridden by passing an integer as the third argument to .

$json = '{"@attributes":{"answer":42}}';
$thing = json_decode($json);

echo $thing->{'@attributes'}->answer; //42
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If you have an integer as property see: How to access object properties with names like integers? as reference.

Someone put JSON in your JSON

It's ridiculous but it happens - there's JSON encoded as a string within your JSON. Decode, access the string as usual, decode that, and eventually get to what you need.

$json = '
{
    "type": "donut",
    "name": "Cake",
    "toppings": "[{ \"type\": \"Glazed\" }, { \"type\": \"Maple\" }]"
}';

$yummy = json_decode($json);
$toppings = json_decode($yummy->toppings);

echo $toppings[0]->type; //Glazed
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Data doesn't fit in memory

If your JSON is too large for{-code-1} to handle at once things start to get tricky. See:

• Processing large JSON files in PHP
• How to properly iterate through a big json file

How to sort it

See: Reference: all basic ways to sort arrays and data in PHP.

Solution:

You can use json_decode() to convert a json string to a PHP object/array.

Eg.

Input:

$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';

var_dump(json_decode($json));
var_dump(json_decode($json, true));
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Output:

object(stdClass)#1 (5) {
    ["a"] => int(1)
    ["b"] => int(2)
    ["c"] => int(3)
    ["d"] => int(4)
    ["e"] => int(5)
}

array(5) {
    ["a"] => int(1)
    ["b"] => int(2)
    ["c"] => int(3)
    ["d"] => int(4)
    ["e"] => int(5)
}
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Few Points to remember:

• json_decode requires the string to be a validjson else it will returnNULL.
• In the event of a failure to decode, can be used to determine the exact nature of the error.
• Make sure you pass inutf8 content, orjson_decode may error out and just return aNULL value.

Solution:

// Using json as php array 

$json = '[{"user_id":"1","user_name":"Sayeed Amin","time":"2019-11-06 13:21:26"}]';

//or use from file
//$json = file_get_contents('results.json');

$someArray = json_decode($json, true);

foreach ($someArray as $key => $value) {
    echo $value["user_id"] . ", " . $value["user_name"] . ", " . $value["time"] . "<br>";
}
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Solution:

We can decode json string into array using json_decode function in php

1) json_decode($json_string) // it returns object

2) json_decode($json_string,true) // it returns array

$json_string = '{
    "type": "donut",
    "name": "Cake",
    "toppings": [
        { "id": "5002", "type": "Glazed" },
        { "id": "5006", "type": "Chocolate with Sprinkles" },
        { "id": "5004", "type": "Maple" }
    ]
}';
$array = json_decode($json_string,true);

echo $array['type']; //it gives donut
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Solution:

Consider usingJSONPath https://packagist.org/packages/flow/jsonpath

There is a pretty clear explanation of how to use it and parse a JSON-file avoiding all the loops proposed. If you are familiar withXPath forXML you will start loving this approach.

Solution:

The acepted Answer is very detailed and correct in most of the cases.

I just want to add that i was getting an error while attempting to load a JSON text file encoded with UTF8, i had a well formatted JSON but the 'json_decode' always returned me with NULL, it was due the BOM mark.

To solve it, i made this PHP function:

function load_utf8_file($filePath)
{
    $response = null;
    try
    {
        if (file_exists($filePath)) {
            $text = file_get_contents($filePath);
            $response = preg_replace("/^\xEF\xBB\xBF/", '', $text);          
        }     
    } catch (Exception $e) {
      echo 'ERROR: ',  $e->getMessage(), "\n";
   }
   finally{  }
   return $response;
}
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Then i use it like this to load a JSON file and get a value from it:

$str = load_utf8_file('appconfig.json'); 
$json = json_decode($str, true); 
//print_r($json);
echo $json['prod']['deploy']['hostname'];
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Solution:

https://paiza.io/projects/X1QjjBkA8mDo6oVh-J_63w

Check below code for converting json to array inPHP, If JSON is correct thenjson_decode() works well, and will return an array, But if malformed JSON, then It will returnNULL,

<?php
function jsonDecode1($json){
    $arr = json_decode($json, true);
    return $arr;
}

// In case of malformed JSON, it will return NULL
var_dump( jsonDecode1($json) );
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If malformed JSON, and you are expecting only array, then you can use this function,

<?php
function jsonDecode2($json){
    $arr = (array) json_decode($json, true);
    return $arr;
}

// In case of malformed JSON, it will return an empty array()
var_dump( jsonDecode2($json) );
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If malformed JSON, and you want to stop code execution, then you can use this function,

<?php
function jsonDecode3($json){
    $arr = (array) json_decode($json, true);

    if(empty(json_last_error())){
        return $arr;
    }
    else{
        throw new ErrorException( json_last_error_msg() );
    }
}

// In case of malformed JSON, Fatal error will be generated
var_dump( jsonDecode3($json) );
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You can use any function depends on your requirement,