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JSONオブジェクトをループする方法

CodeIgniterを使用して、次のJSONの名前をトラバースして表示するにはどうすればよいですか？

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');

class Search extends CI_Controller {
    public function index()
    {       

        $json = '[{"name": "John Doe",
                 "address": "115 Dell Avenue, Somewhere",
                 "tel": "999-3000",
                 "occupation" : "Clerk"},
                 {"name": "Jane Doe",
                 "address": "19 Some Road, Somecity",
                 "tel": "332-3449",
                 "occupation": "Student"}]';


        for (int $i = 0; $i < $json.length; $i++){
            ???
        }
        //$obj = json_decode($json);        
        //$this->load->view('search_page');
    }
}

/* End of file search.php */
/* Location: ./application/controllers/search.php */
phpjsoncodeigniter
22
Anthony

1）$jsonは、最初にデコードする必要がある文字列です。

$json = json_decode($json);

2）オブジェクトをループして、そのメンバーを取得する必要があります

foreach($json as $obj){
   echo $obj->name;
   .....

}
51
Josh

もう一つの例：

  <?php

  //lets make up some data:
  $udata['user'] = "mitch";
  $udata['date'] = "2006-10-19";
  $udata['accnt'] = "EDGERS";
  $udata['data'] = $udata; //array inside
  var_dump($udata); //show what we made

  //lets put that in a file
  $json = file_get_contents('file.json');
  $data = json_decode($json);
  $data[] = $udata;
  file_put_contents('file.json', json_encode($data));

  //lets get our json data
  $json = file_get_contents('file.json');
  $data = json_decode($json);
  foreach ($data as $obj) {
        var_dump($obj->user);
  }
3
Mike Q