如何使用CodeIgniter遍历并显示以下JSON中的名称?
<?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 */
答案 0 :(得分:48)
1)$json
是您需要首先对其进行解码的字符串。
$json = json_decode($json);
2)你需要遍历对象并获得其成员
foreach($json as $obj){
echo $obj->name;
.....
}
答案 1 :(得分:3)
另一个例子:
<?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);
}