停止执行脚本？(Stop executing a script?)

 我从ajax（addContent）调用php函数： 
 
protected $output = array('success'=>0, 'message'=>'There was an error, please try again.');

public function addContent()
{
    $imgName = $this->doSomething();
    $this->doSomethingElse();

    //save imageName to DB
    $this->output['success'] = 1;
    return $output;
}

private function doSomething()
{
    if($imageOk){
       return $imageName;
    }
    else
    {
       $this->output['message'] = 'bad response';
       //how to return output?
    }

}
 
 我已经将这些方法简化为说明目的。 
 
 如果方法'doSomething（）'的输出响应错误，我怎样才能将它从addContent方法发送回ajax？ 如果输出不好，我想退出脚本而不是继续执行doSomethingElse（）。 

I call a php function from ajax (addContent):
 
protected $output = array('success'=>0, 'message'=>'There was an error, please try again.');

public function addContent()
{
    $imgName = $this->doSomething();
    $this->doSomethingElse();

    //save imageName to DB
    $this->output['success'] = 1;
    return $output;
}

private function doSomething()
{
    if($imageOk){
       return $imageName;
    }
    else
    {
       $this->output['message'] = 'bad response';
       //how to return output?
    }

}
 
I have made the methods simple for illustrative purposes.
 
If the method 'doSomething()' has an output of bad response, how can I send this back from the addContent method to ajax? If the output is bad, I want to exit the script and not continue executing doSomethingElse().

原文：https://stackoverflow.com/questions/19450242