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[3단계 - Tomcat 구현하기] 카피(김상혁) 미션 제출합니다. #610

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merged 22 commits into from
Sep 13, 2024

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tkdgur0906
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안녕하세요 잉크~
3단계 미션 진행 후 제출합니다!
전반적으로 리팩토링만 진행했습니다!

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3단계 리팩토링 잘 해주셨습니다 👍
몇 가지 코멘트 남겼습니다

@@ -30,7 +41,7 @@ public static HttpRequest from(InputStream inputStream) throws IOException {
String startLine = bufferedReader.readLine();
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RequestLine 을 분리하면 어떨까요 (lms 3단계 - 힌트 1 참고)

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저런 힌트가 있었군요! HttpRequestLine 클래스를 생성해서 역할을 분리했습니다!

@@ -30,7 +41,7 @@ public static HttpRequest from(InputStream inputStream) throws IOException {
String startLine = bufferedReader.readLine();

String[] parsedStartLine = HttpRequestParser.parseStartLine(startLine);
String method = parsedStartLine[0];
HttpMethod method = HttpMethod.from(parsedStartLine[0]);
String path = parsedStartLine[1];
String version = parsedStartLine[2];

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49~56번줄 while문 리팩토링 해주시면 좋을것 같아요.
지금은 while문의 조건이 true나 다름없습니다!

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현재 while 방식으로는 조건문이 의미가 없고 무조건 if에서 처리되겠네요!
while문의 탈출 조건을 변경하였습니다!

public byte[] toHttpResponseBody() throws URISyntaxException, IOException {
URL url = getClass().getClassLoader().getResource(loadResourceName(path));
if (url == null) {
throw new IllegalArgumentException("존재하지 않는 리소스 입니다." + path);
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404도 고려해주세요

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예외 종류별로 반환코드를 설정하면 좋을 것 같습니다!
다만 다른 기능의 우선순위가 높을 것 같아 추후 고려해보겠습니다!

} catch (IOException | UncheckedServletException e) {
log.error(e.getMessage(), e);
} catch (URISyntaxException e) {
} catch (Exception e) {
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알 수 없는 예외가 발생했을 때 어떤 응답을 주는게 좋을까요?

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알 수 없는 예외가 발생했을 때 500 에러를 반환하는 방어 로직을 작성해야 할 거 같습니다!
이 기능도 위와 동일하게 추후 고려해보겠습니다~

@yechop yechop merged commit 9fa3dd7 into woowacourse:tkdgur0906 Sep 13, 2024
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2 participants