java - 冒泡异常 : Java

Question

这个问题已经有答案了:

How do you implement a re-try-catch? (29 个回答)

已关闭 4 年前。

所以我有一个方法，如果发生异常，我想重试该方法内的操作。如果异常第二次发生，我希望在另一个类调用该方法的地方捕获异常。这是正确的方法吗？

    public OAuth2AccessToken getAccessTokenWithRefreshToken  (String refreshToken) throws OAuth2AccessTokenErrorResponse, IOException, InterruptedException ,ExecutionException  {
    try {
        System.out.println("trying for the first time");
        OAuth2AccessToken mAccessToken = mOAuthService.refreshAccessToken(refreshToken);
        return mAccessToken;
     catch (IOException | InterruptedException | ExecutionException e) {
        try {
            System.out.println("trying for the second time");
            OAuth2AccessToken mAccessToken = mOAuthService.refreshAccessToken(refreshToken);
        }  catch (IOException | InterruptedException | ExecutionException e2) {
            // TODO Auto-generated catch block
            e2.printStackTrace();
            throw e2;
        }
    }
    return mAccessToken;
}

Answer 1

最好使用循环，以免重复:

public OAuth2AccessToken getAccessTokenWithRefreshToken  (String refreshToken) throws OAuth2AccessTokenErrorResponse, IOException, InterruptedException ,ExecutionException {
    int maxAttempts = 2;
    int attempt = 0;
    while (attempt < maxAttempts) {
        try {
            return mOAuthService.refreshAccessToken(refreshToken);
        }
        catch (IOException | InterruptedException | ExecutionException e) {
            attempt++;
            if (attempt >= maxAttempts) {
                throw e;
            }
        }
    }
    return null; // or throw an exception - should never be reached
}