Php Json未定义错误 我有要编辑的操作，然后发送json数据

我想输入该有效值，但当我发出警报时，它仅显示undefined

BtnEdit : function(val){
    var form=$('#edit_salary');
    $.ajax({
        type:'POST',
        data: form.serialize(),
        dataType:'json',
        url : "btnedit",
        success:function(d){
            alert(d.valid);      //undefined
            alert(d);            //{"valid":true}                     
        }
    });
}

用于变换对象中的字符串：

success:function(d){
    d = JSON.parse(d);
    alert(d.valid);      // true
}

---

您需要使用

JSON.parse

来转换对象中的字符串：

success:function(d){
    d = JSON.parse(d);
    alert(d.valid);      // true
}

试一试

---

由于您正在使用Phalcon框架，我建议您使用它的内置功能来正确处理json响应：

$this->response->setContentType('application/json', 'UTF-8');
$this->response->setJsonContent($responseAsArray);
$this->response->send();

如果响应中包含

application/json

的头，jQuery将开始正确处理它，您不需要

json.parse

您的帧

为了在我的一个应用程序中处理API模块中的json响应，我有一个

ControllerBase

，它使用此事件处理程序扩展了

\Phalcon\Mvc\Controller

：

use \Phalcon\Mvc\Controller;

class ControllerBase extends Controller {

/**
 * Captures method result and tries to make a JSON response out of it.
 * 
 * @param \Phalcon\Mvc\Dispatcher $dispatcher
 * @return \Phalcon\Http\Response
 */
protected function afterExecuteRoute($dispatcher) {

    $content = $dispatcher->getReturnedValue();

    if(is_object($content)) {
        if(is_callable(array($content, 'toArray'))) {
            $content = $content->toArray();
        } else {
            $content = (array) $content;
        }
    }

    $frame = $this->getFrame($content, $dispatcher); // protocol frame creation helper

    $this->response->setContentType('application/json', 'UTF-8');
    switch($frame['code']) {
        case 200:
            $this->response->setStatusCode(200, 'OK');
            break;
        case 400: 
            $this->response->setStatusCode(404, 'Not found');
            break;
        case 500: 
            $this->response->setStatusCode(503, 'Service Unavailable');
            break;
    }

    // clearing potential warnings
    $ob = ob_get_clean();
    if(strlen($ob) > 0) {
        /**
         * @todo some logging of $ob !
         * this will be a dead code if you will make an Error2Exception handler
         */
        echo($ob); die();
    }

    // settinf response content as JSON
    $this->response->setJsonContent($frame);

    return $this->response->send();
}
}

---

首先解析

JSON

。您已经谈到了JSON.Parse，但您没有在示例中使用它？？？@AsimShahzad该示例在我的回答中的

下。当然我可以看到它，但请使用您所说的@Hassan方法
$this->response->setContentType('application/json', 'UTF-8');
$this->response->setJsonContent($responseAsArray);
$this->response->send();

use \Phalcon\Mvc\Controller;

class ControllerBase extends Controller {

/**
 * Captures method result and tries to make a JSON response out of it.
 * 
 * @param \Phalcon\Mvc\Dispatcher $dispatcher
 * @return \Phalcon\Http\Response
 */
protected function afterExecuteRoute($dispatcher) {

    $content = $dispatcher->getReturnedValue();

    if(is_object($content)) {
        if(is_callable(array($content, 'toArray'))) {
            $content = $content->toArray();
        } else {
            $content = (array) $content;
        }
    }

    $frame = $this->getFrame($content, $dispatcher); // protocol frame creation helper

    $this->response->setContentType('application/json', 'UTF-8');
    switch($frame['code']) {
        case 200:
            $this->response->setStatusCode(200, 'OK');
            break;
        case 400: 
            $this->response->setStatusCode(404, 'Not found');
            break;
        case 500: 
            $this->response->setStatusCode(503, 'Service Unavailable');
            break;
    }

    // clearing potential warnings
    $ob = ob_get_clean();
    if(strlen($ob) > 0) {
        /**
         * @todo some logging of $ob !
         * this will be a dead code if you will make an Error2Exception handler
         */
        echo($ob); die();
    }

    // settinf response content as JSON
    $this->response->setJsonContent($frame);

    return $this->response->send();
}
}