已知資料：提供資料
1.List<Map<String, String>> inputData
2.List<某物體類> filter，包含一定資料，該物體類包括id和name兩個欄位
3.第一條的key對應id，value對應name
要求篩選出filter中有而inputData中沒有的資料，用stream流和lambda運算式操作
uj5u.com熱心網友回復：
給你段sample代碼吧public class Sample {
static class SomeEntity {
String id;
String name;
public SomeEntity(String id, String name) {this.id=id; this.name=name;}
public String toString() {return String.format("[id:%s, name:%s]", id, name);}
}
public static void main(String[] args) {
List<Map<String, String>> inputData = new ArrayList<>();
for (int i=0; i<10; i+=2) { //id=0，2，4，6，8
Map<String, String> m = new HashMap<>();
m.put("id-"+i, "name-"+i);
inputData.add(m);
}
List<SomeEntity> filter = new ArrayList<>();
for (int i=0; i<5; i++) { //id=0，1，2，3，4
filter.add(new SomeEntity("id-"+i, "name-" + i));
}
for (SomeEntity e : filter1(inputData, filter)) { //從filter中篩選出1，3
System.out.println(e);
}
for (Map<String, String> m : filter2(inputData, filter)) { //從inputData中篩選出6，8
System.out.println(m);
}
}
public static List<SomeEntity> filter1(List<Map<String, String>> inputData, List<SomeEntity> filter) {
return filter.stream().filter(e->
(inputData.stream().filter(in->
(in.containsKey(e.id) && in.get(e.id).equals(e.name))).count()==0)
).collect(Collectors.toList());
}
public static List<Map<String, String>> filter2(List<Map<String, String>> inputData, List<SomeEntity> filter) {
return inputData.stream().filter(in->
(filter.stream().filter(e->
(in.containsKey(e.id) && in.get(e.id).equals(e.name))).count()==0)
).collect(Collectors.toList());
}
}轉載請註明出處，本文鏈接：https://www.uj5u.com/houduan/201634.html
標籤：Java EE
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