【发布时间】:2014-03-29 03:39:59
【问题描述】:
List<String> uniqueidslist = new ArrayList<String>();
uniqueidslist.size() is 7
sample uniqueidslist data --> uniqueidslist = {u1, u2, u3, u4, u5, u6, u7}
List<String> receiveruserfullnamelist= new ArrayList<String>();
receiveruserfullnamelist.size() is 7
sample receiveruserfullnamelist data --> receiveruserfullnamelist = {username1, username2, username3, username4, username5, username6, username7}
我已经创建了一个这样的哈希图:
HashMap<List<String>, List<String>> usermap = new HashMap<List<String>, List<String>>();
usermap.put(uniqueidslist, receiveruserfullnamelist);
我还有另一个这样的 ArrayList:
List<String> finaluserreceiverids = new ArrayList<String>();
finaluserreceiverids.size() is 51
sample finaluserreceiverids data --> finaluserreceiverids = {u1, u4, u7, u1, u1, u1, u2, u4, u5, u2, u2, u2, u3, .....}
现在,我需要一个 ArrayList,其中包含与确切索引相对应的 finaluserreceiverids 中每个 id 的用户名。 像这样:
List<String> finaluserreceivernames = new ArrayList<String>();
finaluserreceivernames.size() is 51
sample finaluserreceivernames data --> finaluserreceivernames = {username1, username4, username7, username1, username1, username1, username2, username4, username5, username2, username2, username2, username3, .....}
我从各种数据库表中动态获取所有这些数据,因此我无法手动将名称放入相应的索引中。
【问题讨论】:
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我认为你的hashmap在这种情况下是错误的,应该是HashMap
,你应该通过迭代 把值放在那里uniqueidslist
标签: android list arraylist hashmap dynamic-data