JSON PHP服务器已修改

Question

The following is my JSON data.

[
  {
    "id": 1,
    "Name": {
      "FirstName": "Wilmer",
      "LastName": "Crona",
      "FullName": "Mr. Cameron Prosacco"
    },
    "Address": {
      "Address1": "84154 Vickie Burg Apt. 994",
      "Address2": "Suite 339",
      "ZipCode": "89080-0376"
    },
    "Phone": "316-269-7694 x1364"
  },
  {
    "id": 2,
    "Name": {
      "FirstName": "Mercedes",
      "LastName": "Kshlerin",
      "FullName": "Dr. Kellie Bashirian"
    },
    "Address": {
      "Address1": "12638 Cali Spurs",
      "Address2": "Suite 353",
      "ZipCode": "76622"
    },
    "Phone": "319-329-3169 x8848"
  }
]

Here is where I want my JSON like this: https://github.com/Zlob/php-json-server but in the method. The JSON file must one file and process only one JSON (db.json).

If my JSON file json1.json , json2.json, etc. How should I do?

Answer 1

You can't decode only first element, you can decode all and get first element only, here is online compiler

<?php

$json = '[
{
    "id": 1,
     "Name": {
     "FirstName": "Wilmer",
     "LastName": "Crona",
     "FullName": "Mr. Cameron Prosacco"
   },
   "Address": {
     "Address1": "84154 Vickie Burg Apt. 994",
     "Address2": "Suite 339",
     "ZipCode": "89080-0376"
   },
   "Phone": "316-269-7694 x1364"
 },
 {
   "id": 2,
   "Name": {
     "FirstName": "Mercedes",
     "LastName": "Kshlerin",
     "FullName": "Dr. Kellie Bashirian"
   },
   "Address": {
     "Address1": "12638 Cali Spurs",
     "Address2": "Suite 353",
     "ZipCode": "76622"
   },
   "Phone": "319-329-3169 x8848"
 }
]';
$data = json_decode($json);
$firstObject = $data[0];
var_dump($firstObject);