How to store the value of the indexes of this JSON in variable?

Question

I have a function in php that returns it to me.

[{"id":81,"username":"usuarioteste23"}]

How can I save in a variable this ID and in another variable this Username?

Thank you.

Answer 1

Solution

Do so:

$dados = json_decode('[{"id":81,"username":"usuarioteste23"}]', true);

$id = $dados[0]['id'];

$username = $dados[0]['username'];

Explanation

What its function returns is a php string, containing serialized data in format JSON.

JSON is used for you to exchange computer data. It uses javascript object notation (no javascript is required to read it). You can easily read it in multiple languages, such as PHP, java or python.

In PHP, the function that decodes serialized data in JSON is json_decode. It will transform a data set of JSON for PHP (array, object, int, etc....).

In the example, I ethylized $dados[0] because his json has an object inside an array. So, we need to take the index 0 to capture the first (and only) value of this array, which in turn contains an object with the id and username properties.

To make it easier to understand (given that you know that JSON is a type of notation similar to javascript types).

The array no json is repeseated by [].

The objeto in json is represented by {}

json_decode return would result in the following statement, if made in php.

$dados = array();

// Object converte para stdClass

$dados[0] = (object)array('id' => 81, 'username' => 'usuarioteste23');

As you can see, I used a second argument passed for the json_decode, which is the value true, que serve para definir se vamos decodificar o json paraarrayoustdClass. Nesse caso trabalhei usando umarray`.

The return of this json_decode with the second argument being true, would be something like that statement:

 $dados = array();

// Sem conversão para stdClass, usei um array simples

$dados[0] = array('id' => 81, 'username' => 'usuarioteste23');

Answer 2

Although the answer has already been given - and chosen - I think it valid to post another solution that is very elegant and that the language brings natively (it seems to have even been built for your case):

$json = '[{"id":81,"username":"usuarioteste23"}]';
# agora você tem $id e $username
extract(array_pop(json_decode($json, true)));

Additional information:

The above code is not incorrect (as stated by someone in the comment below), but it will generate a WARNING (not error) if Strict Standards is enabled:

PHP Strict Standards:  Only variables should be passed by reference

This is because, by default, PHP tries to pass the array_pop|array_shift argument by reference and in this case, should be a variable, not an array.

Note:
There is no relation to bad practice or bad idea - at least in this case - in using the built-in function extract(), since what you want get is a variable $id and another call $username. If you are afraid of collision of variable names, yourself spend a prefix for the third parameter of Extract(). The use of Extract() is popularly used in frameworks (like Laravel) and template Engines.

Having said that, the code should be better written that way:

<?php

$data = json_decode('[{"id":81,"username":"usuarioteste23"}]', true);

extract(array_pop($data));