话不多说,看代码和效果
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/** * 根据map中的某个key 去除list中重复的map * @author shijing * @param list * @param mapkey * @return */ public static list<map<string, object>> removerepeatmapbykey(list<map<string, object>> list, string mapkey){ if (collectionutils.isnullorempty(list)) return null ; //把list中的数据转换成msp,去掉同一id值多余数据,保留查找到第一个id值对应的数据 list<map<string, object>> listmap = new arraylist<>(); map<string, map> msp = new hashmap<>(); for ( int i = list.size()- 1 ; i>= 0 ; i--){ map map = list.get(i); string id = (string)map.get(mapkey); map.remove(mapkey); msp.put(id, map); } //把msp再转换成list,就会得到根据某一字段去掉重复的数据的list<map> set<string> mspkey = msp.keyset(); for (string key: mspkey){ map newmap = msp.get(key); newmap.put(mapkey, key); listmap.add(newmap); } return listmap; } |
测试:
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public static void main(string[] args) { map<string, map> msp = new hashmap<string, map>(); list<map<string, object>> list = new arraylist<map<string, object>>(); list<map<string, object>> listmap = new arraylist<map<string,object>>(); map<string, object> map1 = new hashmap<string, object>(); map1.put( "id" , "1123" ); map1.put( "name" , "张三" ); map<string, object> map2 = new hashmap<string, object>(); map2.put( "id" , "2" ); map2.put( "name" , "李四" ); map<string, object> map3 = new hashmap<string, object>(); map3.put( "id" , "1123" ); map3.put( "name" , "王五" ); map<string, object> map4 = new hashmap<string, object>(); map4.put( "id" , "3" ); map4.put( "name" , "赵六" ); list.add(map1); list.add(map2); list.add(map3); list.add(map4); system.out.println( "初始数据:" + list.tostring()); system.out.println( "去重之后:" + removerepeatmapbykey(list, "id" )); } |
结果:
初始数据:[{name=张三, id=1123}, {name=李四, id=2}, {name=王五, id=1123}, {name=赵六, id=3}]
去重之后:[{name=李四, id=2}, {name=赵六, id=3}, {name=张三, id=1123}]
总结
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原文链接:https://blog.csdn.net/moneyshi/article/details/81220421