獲取json數據到codeigniter腳本的外部

Question

我正在使用codeigniter 3.1.1。我正在嘗試將json數據發送到codeigniter腳本的外部。

笨Valid.php

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

class Valid extends CI_Controller 
{ 

    public function __construct() 
    { 
     parent::__construct(); 
     $this->load->model("user_model"); 

    } 

    public function index() 
    { 

    if($this->user->loggedin) { 
     $json['success'] = "false"; 
    } else { 
     $json['success'] = "true"; 
    } 


    echo json_encode($json); 

    } 

} 

笨Check.php

$url = 'http://url.com/valid'; 

    $curl_handle = curl_init(); 
    curl_setopt($curl_handle,CURLOPT_URL,$url); 
    curl_setopt($curl_handle,CURLOPT_RETURNTRANSFER,TRUE); 
    curl_setopt($curl_handle, CURLOPT_HEADER, FALSE); 
    curl_setopt ($curl_handle,CURLOPT_CONNECTTIMEOUT,0); 
    curl_setopt($curl_handle, CURLOPT_FOLLOWLOCATION, TRUE); 
    $json = curl_exec($curl_handle); 
    curl_close($curl_handle); 

    $json = json_decode($json); 
    echo $json['success']; 

之外我得到$json['success'] = false每次。

Answer 1

嘗試這樣的..

1.Here $this->user->loggedin這是不correct.I認爲需要打電話給你user_model。所以logged_in()功能

public function index() 
    { 
    $this->load->library('user');//loads user library 
    if($this->user->loggedin()) { 
     $json['success'] = "false"; 
    } else { 
     $json['success'] = "true"; 
    } 

2.And在Check.php。你必須有通過true參數與json_decode()轉換成數組。

$json = json_decode($json,true);//converts into array 
echo $json['success'];