对List<Map>里的map的某个属性重复的值进行处理的方法

package test;

import java.util.*;
import java.util.stream.Collectors;

public class Test5 {

public static void main(String[] args) {

List<LinkedHashMap<String, Object>> list = new ArrayList<>();
list.add(new LinkedHashMap<String, Object>() {{
put("id", "a");
put("age", "13");
put("sex", "男");
}});
list.add(new LinkedHashMap<String, Object>() {{
put("id", "a");
put("age", "14");
put("sex", "男");
}});
list.add(new LinkedHashMap<String, Object>() {{
put("id", "b");
put("age", "14");
put("sex", "男");
}});
list.add(new LinkedHashMap<String, Object>() {{
put("id", "c");
put("age", "14");
put("sex", "男");
}});
list.add(new LinkedHashMap<String, Object>() {{
put("id", "");
put("age", "14");
put("sex", "男");
}});
list.add(new LinkedHashMap<String, Object>() {{
put("id", "c");
put("age", "14");
put("sex", "男");
}});


handleRepeatPrimaryKey(list, "id");
List<LinkedHashMap<String, Object>> hasList = new ArrayList<>();
hasList = list.stream().filter(x -> !x.get("id").toString().equals("")).collect(Collectors.toList());
List<LinkedHashMap<String, Object>> notHasList = new ArrayList<>();
notHasList = list.stream().filter(x -> x.get("id").toString().equals("")).collect(Collectors.toList());

System.out.println(hasList);
System.out.println(notHasList);


}


public static List<LinkedHashMap<String, Object>> handleRepeatPrimaryKey(List<LinkedHashMap<String, Object>> list, String id) {
List<LinkedHashMap<String, Object>> itemList = new ArrayList<>();
for (LinkedHashMap<String, Object> lhm : list) {
if (itemList.size() == 0) {
itemList.add(lhm);
} else {
ListIterator<LinkedHashMap<String, Object>> listIterator = itemList.listIterator();
while (listIterator.hasNext()) {
if (listIterator.next().get(id).toString().equals(lhm.get(id).toString())) {
lhm.put(id, "");
}
}
// 注意放在循环外面
listIterator.add(lhm);
}
}
return list;
}
}

对重复主键进行处理,取第一个,剩余的主键都置空,嗯,能满足我的需求。不知道还有没有更好的方法?

posted @ 2019-07-28 23:58  大水煮鱼  阅读(6369)  评论(0编辑  收藏  举报