从“stream（）”或“parallelStream（）”中捕获异常失去了正确的值

本文介绍了从“stream（）”或“parallelStream（）”中捕获异常失去了正确的值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 在下面的代码中，当为迭代捕获 code> stream（）或 blockquote> 解决方案为什么不把lambda-body包装在 您也可以在 map 之后过滤 null $ b $ pre $ intList = strList.stream（）//与parallelStream（）相同 .map（x - > ; { try { return Integer.parseInt（x）; } catch（NumberFormatExce ption nfe）{ System.err.println（nfe.getMessage（））; } 返回null; 。） .filter（x - > x！= null） .collect（Collectors.toList（））; 这会给你想要的 intList = [1,2,4,6编辑：为了减少lamdba中try / catch的沉重感，你可以添加一个辅助方法。 static Integer parseIntOrNull（String s）{ try { return Integer.parseInt（s）; } catch（NumberFormatException nfe）{ System.err.println（nfe.getMessage（））; } 返回null; intList = strList.stream（） .map（x - > parseIntOrNull（x）） .filter（x - > x！= null） .collect（Collectors.toList（））;或者为避免使用null，您可以返回一个Stream $ b static Stream< Integer> parseIntStream（String s）{ try { return Stream.of（Integer.parseInt（s））; } catch（NumberFormatException nfe）{ System.err.println（nfe.getMessage（））; } 返回Stream.empty（）; $ b $ intList = strList.stream（） .flatMap（x - > parseIntStream（x）） .collect（Collectors.toList（））; In the following code, when catching NumberFormatException out of for iteration, the strings in appropriate form appearing in strList before the first bad one (i.e., "illegal_3") have been parsed successfully (i.e., "1" and "2" have been parsed as integers 1 and 2).public void testCaughtRuntimeExceptionOutOfIteration() { List<String> strList = Stream.of("1", "2", "illegal_3", "4", "illegal_5", "6").collect(Collectors.toList()); List<Integer> intList = new ArrayList<>(); try{ for (String str : strList) { intList.add(Integer.parseInt(str)); } } catch (NumberFormatException nfe) { System.err.println(nfe.getMessage()); } List<Integer> expectedIntList = Stream.of(1, 2).collect(Collectors.toList()); // passed assertEquals("The first two elements have been parsed successfully.", expectedIntList, intList); }However, when replacing for iteration by stream() or parallelStream(), I lose 1 and 2.public void testCaughtRuntimeExceptionOutOfStream() { List<String> strList = Stream.of("1", "2", "illegal_3", "4", "illegal_5", "6").collect(Collectors.toList()); List<Integer> intList = new ArrayList<>(); try{ intList = strList.stream() // same with "parallelStream()" .map(Integer::parseInt) .collect(Collectors.toList()); } catch (NumberFormatException nfe) { System.err.println(nfe.getMessage()); } List<Integer> expectedIntList = Stream.of(1, 2).collect(Collectors.toList()); // failed: expected:<[1,2]>, but was:<[]> assertEquals("The first two elements have been parsed successfully.", expectedIntList, intList); } What is the specification of the control flow of exceptions thrown from within stream() or parallelStream()? How can I get the result of intList = [1,2] (i.e., ignore the ones after the first NumberFormatException is thrown) or even better intList = [1,2,4,6] (i.e., ignore the bad ones with NumberFormatException) with stream() or parallelStream() 解决方案 Why not just wrap lambda-body in try...catch?Also you can filter null values after map: intList = strList.stream()// same with "parallelStream()" .map(x -> { try { return Integer.parseInt(x); } catch (NumberFormatException nfe) { System.err.println(nfe.getMessage()); } return null; }) .filter(x -> x!= null) .collect(Collectors.toList());This will give you desired intList = [1,2,4,6].Edit: To reduce the "heaviness" of a try/catch in a lamdba you can add a helper method.static Integer parseIntOrNull(String s) { try { return Integer.parseInt(s); } catch (NumberFormatException nfe) { System.err.println(nfe.getMessage()); } return null;}intList = strList.stream() .map(x -> parseIntOrNull(x)) .filter(x -> x!= null) .collect(Collectors.toList());Or to avoid using null, you can return a Streamstatic Stream<Integer> parseIntStream(String s) { try { return Stream.of(Integer.parseInt(s)); } catch (NumberFormatException nfe) { System.err.println(nfe.getMessage()); } return Stream.empty();}intList = strList.stream() .flatMap(x -> parseIntStream(x)) .collect(Collectors.toList()); 这篇关于从“stream（）”或“parallelStream（）”中捕获异常失去了正确的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！