使用PHP解码嵌套的JSON字符串

Question

**我想转换json响应并显示值**

<?php
// json string
$json_string = '{
    "message": "Successfully retrived",
    "error": false,
    "result": [{
        "id": 6128071,
        "name": "jhon doe",
        "number": "31231231230",
        "city": "C",
        "cnic": "8982378237897278",
        "address": "address123",
        "activation_date": "0"
    }]
}';

// decode json string
$json_array = json_decode($json_string, true);


?>

// Print Values
Name : <?=$json_array['result']['name']?><br>
Number: <?=$json_array['result']['number']?><br>
Cnic: <?=$json_array['result']['cnic']?><br>
Address : <?=$json_array['result']['address']?><br>
activation date: <?=$json_array['result']['activation_date']?>

但是我在这条线上出错了

$ json_array ['result'] ['name'];

注意：未定义索引：名称

请指导我如何解决此错误。我尝试搜索Google，但未找到任何解决方案。