使用Doctrine \ DBAL \ Exception捕获db错误

I want to catch any error (ie: foreign key error) from an insert statement or any other. How can I achive that with use Doctrine\DBAL\Exception?

我想从insert语句或任何其他语句中捕获任何错误（即：外键错误）。如何使用Doctrine \ DBAL \ Exception来实现？

I have this when I do the insert:

插入时我有这个：

            $db->beginTransaction();
            try {
              $db->insert('bbc5.produccionpry',$datos);
              $datos['propryid'] = $db->lastInsertId();
              $db->commit();
              // $db = null;
              $resp[] = $datos;
            } catch (Exception $e) {
              $error = array_merge($error, array('error' => $e->errorInfo()));
              $db->rollback();
              throw $e;
            }

But, that doesn't prevent the concrete5 to return a website telling about the error, so, I don't want that website to be shown, I want to catch the error in an array() in order to return it via echo json_encode($error)

但是，这并不妨碍concrete5返回一个告诉错误的网站，所以，我不希望显示该网站，我想在数组（）中捕获错误，以便通过echo json_encode返回它（$错误）

I'm not using the controller for a page, I'm using it for managing RESTful calls from my JavaScript App with this code:

我没有将控制器用于页面，我使用它来管理来自我的JavaScript应用程序的RESTful调用，使用以下代码：

return fetch(`/scamp/index.php/batchprodpry/${maq}`, { 
  method: 'POST', 
  credentials: 'same-origin', 
  headers: { 
    'Accept': 'application/json', 
    'Content-Type': 'application/json' 
  }, 
  body: JSON.stringify(this.state.a) 
})

I'm using ReactJS

我正在使用ReactJS

Thank you

谢谢

2 个解决方案

#1

Don't throw the exception.

不要抛出异常。

Instead of throwing the exception, just get the exceptions message $e->getMessage() from the DBALException $e object and encode it as a JSON string. Important: Put an exit; after the echo to assure that no further code is executed.

不要抛出异常，只需从DBALException $ e对象中获取异常消息$ e-> getMessage（）并将其编码为JSON字符串。重要提示：退出;在echo之后确保没有执行进一步的代码。

use Doctrine\DBAL\DBALException;

try {
    $db->insert('bbc5.produccionpry',$datos);
    $datos['propryid'] = $db->lastInsertId();
    $db->commit();
    $resp[] = $datos;
} 
catch(DBALException $e){
    $db->rollback();
    echo \Core::make('helper/json')->encode($e->getMessage());
    exit;
}

If this code is inside a page controller you could do this:

如果此代码在页面控制器内，您可以这样做：

try {
    $db->insert('bbc5.produccionpry',$datos);
    $datos['propryid'] = $db->lastInsertId();
    $db->commit();
    $resp[] = $datos;
}
catch(DBALException $e){
    $this->error->add($e->getMessage());
}

if($this->error->has()) {
    // All variables that have been set in the view() method will be set again.
    // That is why we call the view method again
    $this->view();
    return;
}

And concrete5 will take care of displaying an appropriate error message.

而concrete5将负责显示相应的错误消息。

Or you could save the $e->getMesssage() in the session and call it inside the view:

或者你可以在会话中保存$ e-> getMesssage（）并在视图中调用它：

$session = \Core::make('session');
// ...
catch(Exception $e){
    $session->set('error', $e->getMessage());
}

And in the view:

在视图中：

// html
<?php 
$session = \Core::make('session');
if($session->has('error')) {
    $m = $session->get('error');
    ?>
    <div id="session_error" class="alert alert-danger">
        <a href="#" class="close">&times;</a>

        <div id="session_error_msg">
            <?php  print $m ?>
        </div>
    </div>
<?php 
}
$session->remove('error');
?>
//html

#2

Since you have throw $e in your catch block you throw this exception further which means that it is (probably) handled by global exception listener. Once you throw exception you exit your code immediately, so a solution for you is just remove throw $e line and you should be good

由于你在catch块中抛出了$ e，因此你会进一步抛出这个异常，这意味着它（可能）由全局异常监听器处理。一旦你抛出异常就立即退出你的代码，所以一个解决方案就是删除throw $ e行，你应该很好

#1