对列表中的数据进行唯一处理

    技术2022-05-11  78

    HashMap hproviders = new HashMap();         for (int i = 0; i < size; i++) {             PjMaterial item = (PjMaterial) Datas.get(i);             if(item == null) continue;                            Material mat = item.getMaterial();             if(mat == null) continue;             List matprovider = (List)mat.getProviders();             int count = matprovider == null?0:matprovider.size();             for(int j = 0; j < count; j ++) {                 WqproviderUser prov = (WqproviderUser)matprovider.get(j);                 if(hproviders.get(prov.getUserId() + "") == null) {                     hproviders.put(prov.getUserId() + "", "");                                       PjProvider pjp = new PjProvider(prov, proj);                     providers.add(pjp);                 }             }         }

    最新回复(0)