php - Laravel:是否可以在捕获异常时记录堆栈跟踪并继续执行？

Question

Laravel 在异常捕获时具有可读的日志和堆栈跟踪，例如:

production.ERROR: Command "test" is not defined.

Did you mean this?
    make:test {"exception":"[object] (Symfony\\Component\\Console\\Exception\\CommandNotFoundException(code: 0): Command \"test\" is not defined.

Did you mean this?
    make:test at {root}/vendor/symfony/console/Application.php:618)
[stacktrace]
#0 {root}/vendor/symfony/console/Application.php(229): Symfony\\Component\\Console\\Application->find('test')
#1 {root}/vendor/symfony/console/Application.php(148): Symfony\\Component\\Console\\Application->doRun(Object(Symfony\\Component\\Console\\Input\\ArgvInput), Object(Symfony\\Component\\Console\\Output\\ConsoleOutput))
#2 {root}/vendor/laravel/framework/src/Illuminate/Console/Application.php(88): Symfony\\Component\\Console\\Application->run(Object(Symfony\\Component\\Console\\Input\\ArgvInput), Object(Symfony\\Component\\Console\\Output\\ConsoleOutput))
#3 {root}/vendor/laravel/framework/src/Illuminate/Foundation/Console/Kernel.php(121): Illuminate\\Console\\Application->run(Object(Symfony\\Component\\Console\\Input\\ArgvInput), Object(Symfony\\Component\\Console\\Output\\ConsoleOutput))
#4 {root}/artisan(37): Illuminate\\Foundation\\Console\\Kernel->handle(Object(Symfony\\Component\\Console\\Input\\ArgvInput), Object(Symfony\\Component\\Console\\Output\\ConsoleOutput))
#5 {main}
"}

问题是:是否可以自己捕获异常，并记录相同格式的stacktrace，并继续程序执行。到目前为止，我正在通过 Log::error(json_encode(debug_backtrace())); 记录错误这真的很难看，很难追踪。示例代码:

try {
    foo();
} catch(\Exception $e) {
    Log::error(json_encode(debug_backtrace()));
}
bar();

Answer 1

Laravel 有一个专门用于此的辅助方法，请参阅 rescue方法。

return rescue(function () { 
    return $this->method(); 
});

使用您的示例，它看起来像:

rescue(function () { 
    return foo(); 
});

bar();