ThinkPHP 5 中AJAX跨域请求头设置方法

最近用thinkphp做项目，在测试环境时，存在接口的测试问题。在tp官网也没能找到相关的解决方法。自已看了一下源码，有如下的解决方案。

在项目目录下面，创建common/behavior/CronRun.php文件，文件内容如下：

<?php
/**
 * Created by PhpStorm.
 * User: LiuYang
 * Date: 2017/3/9
 * Time: 19:37
 */

namespace app\common\behavior;

use think\Exception;
use think\Response;

class CronRun
{
    public function run(&$dispatch){
        $host_name = isset($_SERVER['HTTP_ORIGIN']) ? $_SERVER['HTTP_ORIGIN'] : "*";
        $headers = [
            "Access-Control-Allow-Origin" => $host_name,
            "Access-Control-Allow-Credentials" => 'true',
            "Access-Control-Allow-Headers" => "x-token,x-uid,x-token-check,x-requested-with,content-type,Host"
        ];
        if($dispatch instanceof Response) {
            $dispatch->header($headers);
        } else if($_SERVER['REQUEST_METHOD'] === 'OPTIONS') {
            $dispatch['type'] = 'response';
            $response = new Response('', 200, $headers);
            $dispatch['response'] = $response;
        }
    }
}

接着在项目中（tags.php）配置行为动作，如下：

<?php
// +----------------------------------------------------------------------
// | ThinkPHP [ WE CAN DO IT JUST THINK ]
// +----------------------------------------------------------------------
// | Copyright (c) 2006~2016 http://thinkphp.cn All rights reserved.
// +----------------------------------------------------------------------
// | Licensed ( http://www.apache.org/licenses/LICENSE-2.0 )
// +----------------------------------------------------------------------
// | Author: liu21st <liu21st@gmail.com>
// +----------------------------------------------------------------------

// 应用行为扩展定义文件
return [
    // 应用初始化
    'app_init'     => [],
    // 应用开始
    'app_begin'    => [
        'app\\common\\behavior\\CronRun'
    ],
    // 模块初始化
    'module_init'  => [],
    // 操作开始执行
    'action_begin' => [],
    // 视图内容过滤
    'view_filter'  => [],
    // 日志写入
    'log_write'    => [],
    // 应用结束
    'app_end'      => [
        'app\\common\\behavior\\CronRun'
    ],
];

ok，以上几步就解决跨域请求问题。