Question

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（）。

Answer 1

如果验证失败，只需让doSomething返回false，然后检查该条件：

public function addContent()
{
    $imgName = $this->doSomething();
    if ($imgName === false) {
        return "your error message";
    }
    $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';
       return false;
    }
}

您可以根据评论使用exit或die，但这些都会立即终止脚本，因此除非您在调用这些函数之前回显或将错误消息分配给模板，否则不会返回错误消息。

Answer 2

看到它解决（抛出异常）：

protected $output = array('success'=>0, 'message'=>'There was an error, please try again.');

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

        //save imageName to DB
        $this->output['success'] = 1;
        return $this->output;
    }
    catch(Exception $e){
        $this->output['success'] = 0;
        $this->output['message'] = $e->getMessage();
        return $this->output;
    }
}

private function doSomething()
{
    if($imageOk){
       return $imageName;
    }
    else
    {
       throw new Exception('bad response');
    }

}

Answer 3

您可以使用异常处理。

protected $output = array('success'=>0, 'message'=>'There was an error, please try again.');

public function addContent()
{
    try {
        $imgName = $this->doSomething();
    } catch (Exception $e) {
        $this->output['message'] = $e->getMessage();
        return $this->output;
    }
    $this->doSomethingElse();

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

private function doSomething()
{
    if($imageOk) {
       return $imageName;
    }
    else {
       throw new Exception('bad response');
    }

}

停止执行脚本？

3 个答案: