Question

我是代码点火器的新手，对于我的项目，我必须创建一个REST API。

我有一个有效的REST API。

从ajax调用调用API的正确方法是什么？

View ->REST

或

View -> Controller -> REST

如果是<?php defined('BASEPATH') OR exit('No direct script access allowed'); require APPPATH . '/libraries/REST_Controller.php'; class User extends REST_Controller { function _construct(){ parent::_construct(); $this->load->database(); $this->load->model('usermodel'); } public function user_get(){ $this->load->model('usermodel'); $requestType = $this->get('req');//get method request switch ($requestType) { case 'login': $username = $this->get('un'); $password = $this->get('pw'); $user = $this->usermodel->getUser($username, $password); if ($user){ $this->response($user, REST_Controller::HTTP_OK); }else{ $this->response([ 'status' => FALSE, 'message' => 'No users were found' ], REST_Controller::HTTP_NOT_FOUND); } break; default: break; } } }，如何从控制器调用REST API？

这是我的类，它扩展了用户的rest_controller

    function signIn(){
    var username = $('#signInUn').val();
    var password = $('#signInPw').val();
    $.ajax({
        method: "GET",
        url : "/infoshare/index.php/user/user/req/login/un/"+username+"/pw/"+password,
        dataType: 'JSON',
        success: function(data){
            var response = data.response;
            var status = data.status;
            if(status == 200){
                var userId = response[0].user_id;
                //success of user 
            }else{
                console.log(data);
            }
        },
        error: function(status){
            console.log("Error occured");
        }
    });
}

这是我的登录用户

登录页面的ajex调用

redirect_to 'user#show'

Answer 1

是的，你可以通过ajax调用rest apil。

$.ajax({ 
type: "GET",
data: {id:1123},
url: "http://localhost:8080/restws/json/user/get",
success: function(data){        
 alert(data);
}

}）;

传递方法所需的数据。如果POST或GET，请检查您的请求类型。

从控制器codeigniter调用REST API

1 个答案: