How to decode JSON to PHP array

Question

I have a JSON output I would like to convert to a PHP array. I tried with json_decode(), the problem is that there are arrays in the arrays. They are the first weapons with PHP and I have never used JSON. Can anyone help me?

Here is the JSON code:

{
    "a": "text",
    "b": "",

    "c": [

        {"name": "1", "id": "some text 1", "val": "x"},
        {"name": "2", "id": "some text 2", "val": "x"},
        {"name": "3", "id": "some text 3", "val": "x"}

    ]

}

I have to check that a variable is equal to name 1, contained in c, and if so, it also takes its id and val.

How can I do?

PS: I can compare two variables, but I do not know how to find name 1 and the corresponding data ..

It's OK, you just decode it once, and all will be an array inside array too — Josua M C
– Josua M C, Commented Jun 29, 2017 at 3:00
Loop through the c array and test the value of name. How is this complicated? — Capsule
– Capsule, Commented Jun 29, 2017 at 3:06
Sorry, I never worked with a array inside an array, now understood. Thank you all :) — Takabrycheri
– Takabrycheri, Commented Jun 29, 2017 at 3:11

Pablo Corezzola · Accepted Answer · 2017-06-29 03:04:17Z

1

$json = '{"a":"text","b":"","c":[{"name":"1","id":"some text 1","val":"x"},{"name":"2","id":"some text 2","val":"x"},{"name":"3","id":"some text 3","val":"x"}]}';

$json = json_decode($json,true);

echo $json["a"]."<br>";
echo $json["b"]."<br>";
echo $json["c"][1]["name"]."<br>";
echo "<pre>".print_r($json,true)."</pre>";

answered Jun 29, 2017 at 3:04

Pablo Corezzola

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