怎么把throw的错误用json输出

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技术zhai
技术zhai

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2018-08-08 15:20:57 提问

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public function run(){

    try{

        $this->_setRequestMethod();

        $this->_setupResource();

        $this->runcontroller();

    }catch(Exception $e){

        $this->_json(['error'=>$e->getMessage()],$e->getCode());
    }

}

public function _setRequestMethod(){

    $this->_requestMethod=$_SERVER['REQUEST_METHOD'];

    if(!in_array($this->_requestMethod,$this->_allowRequestMethods)){
        throw new\Exception("请求方法不被允许", 405);

    }

}

public function _json($array,$code=0){

    if($array === null && $code === 0){

        $code = 204;

    }

    if($array !== null && $code === 0){

        $code = 200;

    }

    header("HTTP/1.1 ".$code."  ".$this->_statusCodes[$code]);

    header("Content-Type=application/json;charset=UTF-8 ");

    if($array !== null){

        echo  json_encode($array,JSON_UNESCAPED_UNICODE);

    }

    exit();

}

执行run()的时候返回

Fatal error: Uncaught Exception: 请求方法不被允许 in /www/wwwroot/www.entercode.cn/api/Rest/restful.php:149 Stack trace: #0 /www/wwwroot/www.entercode.cn/api/Rest/restful.php(69): Restrestful->runcontroller() #1 /www/wwwroot/www.entercode.cn/api/index.php(20): Restrestful->run() #2 {main} thrown in /www/wwwroot/www.entercode.cn/api/Rest/restful.php on line 149

但是我想将返回的错误结果
{error':'请求方法不被允许','405'}

是我哪里写错了吗?

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1个回答

  • 半盛&梦夏149声望 2018-08-08 11:44

    class Etest { public function run() { try { $this->A(); $this->B(); $this->C(); } catch(\Exception $e) { $this->_json(['error' => $e->getMessage(),'code' => $e->getCode()]); } } public function A() { } public function B() { throw new \Exception("Not Found!",404); } public function C() { } public function _json($aArray) { header("HTTP/1.1 ".$aArray['code']." ".$aArray['error']); header("Content-Type:application/json;charset=utf-8;"); echo json_encode($aArray,JSON_UNESCAPED_UNICODE); exit; } } (new Etest)->run(); 可以参考下 这样写无报错 可以正常捕获Exception
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