在Framework WebAPI 2中,我有一个如下所示的控制器:
[Route("create-license/{licenseKey}")]
public async Task<LicenseDetails> CreateLicenseAsync(string licenseKey, CreateLicenseRequest license)
{
try
{
// ... controller-y stuff
return await _service.DoSomethingAsync(license).ConfigureAwait(false);
}
catch (Exception e)
{
_logger.Error(e);
const string msg = "Unable to PUT license creation request";
throw new HttpResponseException(HttpStatusCode.InternalServerError, msg);
}
}
果然,我收到了500条错误消息。
如何在ASP.NET Core Web API中执行类似的操作?
HttpRequestException
似乎不存在。我宁愿继续返回对象而不是HttpRequestMessage
。
答案 0 :(得分:13)
这样的事情怎么样?创建一个中间件,您将在其中公开某些异常消息:
public class ExceptionMiddleware
{
private readonly RequestDelegate _next;
public ExceptionMiddleware(RequestDelegate next)
{
_next = next;
}
public async Task Invoke(HttpContext context)
{
try
{
await _next(context);
}
catch (Exception ex)
{
context.Response.ContentType = "text/plain";
context.Response.StatusCode = (int)HttpStatusCode.InternalServerError;
if (ex is ApplicationException)
{
await context.Response.WriteAsync(ex.Message);
}
}
}
}
在您的应用中使用它:
app.UseMiddleware<ExceptionMiddleware>();
app.UseMvc();
然后在你的行动中抛出异常:
[Route("create-license/{licenseKey}")]
public async Task<LicenseDetails> CreateLicenseAsync(string licenseKey, CreateLicenseRequest license)
{
try
{
// ... controller-y stuff
return await _service.DoSomethingAsync(license).ConfigureAwait(false);
}
catch (Exception e)
{
_logger.Error(e);
const string msg = "Unable to PUT license creation request";
throw new ApplicationException(msg);
}
}
更好的方法是返回IActionResult
。这样你就不必抛出异常了。像这样:
[Route("create-license/{licenseKey}")]
public async Task<IActionResult> CreateLicenseAsync(string licenseKey, CreateLicenseRequest license)
{
try
{
// ... controller-y stuff
return Ok(await _service.DoSomethingAsync(license).ConfigureAwait(false));
}
catch (Exception e)
{
_logger.Error(e);
const string msg = "Unable to PUT license creation request";
return StatusCode((int)HttpStatusCode.InternalServerError, msg)
}
}
答案 1 :(得分:5)
最好不要在每个动作中捕获所有异常。只需捕获您需要特别做出反应的异常并捕获(并包装到HttpResponse)Middleware中的所有其余内容。