Java 8流API-基于条件过滤并收集对象
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Java 8 stream API- Filter based on condition and collect the object
提问by Learn Hadoop
In java 8 , collect emp object based on some filter condition.
在 java 8 中，根据一些过滤条件收集 emp 对象。
in main class
在主班
List<Emp> empList = Arrays.asList(
new Emp("aaa", language1),
new Emp("cc", language2),
new Emp("bb", language3),
new Emp("dd", language3)
);
empList.stream()
.flatMap(s->s.getLanguage().stream())
.filter(s->s.equals("java"))
.forEach(System.out::println); //Here just i am printing.
Actually I need to collect new List<EMP>
其实我需要收集 new List<EMP>
How to collect emp object who are all have language "java". How can I do?
如何收集都有语言“java”的emp对象。我能怎么做？
回答by Sweeper
You should not use flatMapif you want to collect Empobjects in the end because it will change every element to something else and it can be quite hard to map them back.
flatMap如果您Emp最终想收集对象，则不应使用，因为它会将每个元素更改为其他元素，并且很难将它们映射回来。
You should put all your logic in a filter: "keep the Empobject if getLanguagecontains "java"".
您应该将所有逻辑放在filter：“Emp如果getLanguage包含，则保留对象"java"”。
empList.stream()
.filter(x->x.getLanguage().contains("java"))
.collect(Collectors.toList());
回答by Ravi Sapariya
You can also do like this.
你也可以这样做。
List<Object> optionMetas = new ArrayList<>();
Map<Long, Object> optionIdMetaMap_ = optionMetas.stream().filter(option -> option.getXX() || option.getXXX().equal("java"))
.collect(Collectors.toMap(Object::getKEY, item-> item));
Add your relevant condition in filter()
添加您的相关条件 filter()
