如何将延迟参数化“退避弹簧引导”重试
如何将延迟参数化“退避弹簧引导”重试
提问于 2022-03-22 16:00:02
我不能用我的财产作为标号来拖延：
@Value("${delayReintentos}")
private long delay;
@Retryable(value = { SQLException.class }, maxAttempts = 3, backoff = @Backoff(delay = delay))
public String simpleRetry() throws SQLException {
counter++;
LOGGER.info("Billing Service Failed "+ counter);
throw new SQLException();
}Java11，弹簧引导
回答 2
Stack Overflow用户
发布于 2022-03-22 16:25:28
使用delayExpression代替
/**
* An expression evaluating to the canonical backoff period. Used as an initial value
* in the exponential case, and as a minimum value in the uniform case. Overrides
* {@link #delay()}.
* @return the initial or canonical backoff period in milliseconds.
* @since 1.2
*/
String delayExpression() default "";参见自述文件：https://github.com/spring-projects/spring-retry#readme
您可以使用SpEL或属性占位符。
@Backoff(delayExpression = "${my.delay}",
maxDelayExpression = "@integerFiveBean", multiplierExpression = "${onePointOne}")Stack Overflow用户
发布于 2022-03-22 16:27:19
通过在大括号中包装属性名来尝试这一点。
@Retryable(value = SQLException.class, maxAttempts = 3, backoff = @Backoff(delayExpression = "${delay}")页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/71575196
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