Laravel使用PostgreSQL显示"；`try 读取null`上的属性，而不是SQL错误消息

这应该意味着Laraveltry 访问由于底层SQL错误而尚未实例化的模型实例上的属性.

确保您的Laravel应用程序为debug mode，以获取更详细的错误消息.这可以通过在.env文件中设置APP_DEBUG=true来实现.但是，在生产环境中使用此设置时要谨慎.

判断Laravel中是否启用了logging来捕获所有错误，包括SQL错误.Laravel使用Monolog library，它允许详细的错误记录.

例如，要捕获并记录数据库查询异常:

use Illuminate\Database\QueryException;
use Log;

try {
    $letters = Letter::where('aaa', 'dd')->get();
} catch (QueryException $e) {
    Log::error($e->getMessage());
    // Optionally, return a more user-friendly message or take other appropriate actions
    return response()->json(['error' => 'A database error occurred.'], 500);
}

通过捕捉QueryException，您可以记录详细的错误消息，甚至在响应中返回它以用于调试目的.请注意，在生产环境中公开详细的错误消息可能会带来安全风险.

You can also customize the exception handling in your app/Exceptions/Handler.php file to catch and handle database-related exceptions more transparently.
Modify the render or report methods. That allows you to intercept specific exceptions, such as QueryException, and handle them accordingly.

在Handler.php文件的顶部导入必要的异常类:

use Illuminate\Database\QueryException;
use Symfony\Component\HttpFoundation\Response;

report方法用于记录异常.如果需要，您可以对其进行自定义，以添加特定的数据库异常日志(log).该步骤是可选的，并基于您的日志(log)首选项:

public function report(Throwable $exception)
{
    if ($exception instanceof QueryException) {
        // Log the error or send it to an external service like Sentry or Bugsnag
        Log::error($exception->getMessage());
    }

    parent::report($exception);
}

render方法负责将异常转换为发送回客户端的HTTP响应.

public function render($request, Throwable $exception)
{
    // Check if the exception is an instance of QueryException
    if ($exception instanceof QueryException) {
        // Determine if the app is in debug mode
        if (config('app.debug')) {
            // Return a more detailed error response in debug mode
            $errorResponse = [
                'error' => 'Database query error',
                'message' => $exception->getMessage(),
                'sql' => $exception->getSql(),
                'bindings' => $exception->getBindings(),
            ];
            return response()->json($errorResponse, Response::HTTP_INTERNAL_SERVER_ERROR);
        } else {
            // Return a generic error response in production mode
            return response()->json(['error' => 'A database error occurred. Please try again later.'], Response::HTTP_INTERNAL_SERVER_ERROR);
        }
    }

    // Process other exceptions as usual
    return parent::render($request, $exception);
}

When a QueryException is caught, you can decide how to respond based on whether your application is in debug mode. In debug mode, you could return detailed error information, including the SQL statement and bindings.
In production mode, you would typically return a generic error message to avoid exposing sensitive information.

the comments年度的OP Hosserin Ibrahim份报告:

My problem is solved. The problem is a line in my custom handler.
I have programed the system to report any exception in telegram, if happen any error in handler no way to report this.
Just solve this line and Laravel return query exceptions in Postgres.