Question

你好，我在比较两个数组列表时感到困惑，我的一个数组列表为：

 private ArrayList<String> members = new ArrayList<>();
 members.add("member123keyxyzmember123 number");
 members.add("member456keyxyzmember456 number");
 members.add("member789keyxyzmember789 number");
 members.add("member2233keyxyzmember2233 number");
 members.add("member1122keyxyzmember1122 number");

第二个arraylist是：

 private ArrayList<String> syncMembers = new ArrayList<>();
 syncMembers.add("member123keyxyz123statuskeyxyz123photokeyxyzmember123 number");
 syncMembers.add("member456keyxyz456statuskeyxyz456photokeyxyzmember456 number");

问题是我正在比较两者，以便他们给我成员列表中的数字，而不是 syncMembers 列表！

这就是输出应该是：

    member789 number
    member2233 number
    member1122 number

仅！

我一直在尝试的是：

   for (int i = 0; i < members.size(); i++) {
        String stringFromMembersList = members.get(i);
        String[] memberParts = stringFromMembersList.split("keyxyz");
        String memberNumber = memberParts[1];
        //Log.e("hgax", "sync:::" + memberNumber);

        for (int j = 0; j < syncMembers.size(); j++) {
            String stringFromSyncList = syncMembers.get(j);
            String[] syncParts = stringFromSyncList.split("keyxyz");

            String n = syncParts[3];

            if (memberNumber.equals(n)) {
                //Log.e("hgax", "hee:::" + n);
                break;
            } else {
                Log.e("hgax", "ssshee:::" + memberNumber);
            }
        }
    }

我得到的输出是：

   member456 number
   member789 number
   member789 number
   member2233 number
   member2233 number
   member2233 number
   member1122 number
   member1122 number
   member1122 number
   member1122 number

我有点混淆发生在我身上的事情以及我做错了什么？有人可以告诉我在做什么大错的事先谢谢

Answer 1

修改：更新了答案，以便在arrayList

中包含原始混合值

ArrayList<String> membersList = new ArrayList<>();
        ArrayList<String> syncMembersList = new ArrayList<>();
        for (int i = 0; i < members.size(); i++) {
            String s = members.get(i).substring(members.get(i).lastIndexOf("member"));
            membersList.add(s);
        }
        for (int j = 0; j < syncMembers.size(); j++) {
            String s = syncMembers.get(j).substring(syncMembers.get(j).lastIndexOf("member"));
            syncMembersList.add(s);
        }
        for (int i = 0; i < membersList.size(); i++) {
            if (!syncMembersList.contains(membersList.get(i))) {
                System.out.println(membersList.get(i));
            }
        }

这将打印您需要的输出。

Answer 2

尝试考虑一下您需要检查什么才能达到目标。为了确定members中列表syncMembers中的成员不存在，您必须检查该成员的syncMembers列表的全部内容。由于列表不相同（如您的问题所述），因此您无法使用Collection.contains(Object o)。

这应该达到你的目标：

// We need this initial check as if syncMembers is empty it'll display all members
// And why bother to do this if syncMembers is empty anyway!
if (!syncMembers.isEmpty()) {
    for (String member : members) {
        String memberNo = member.split("keyxyz")[1];
        int i = 0;
        boolean found = false;
        while (!found && i < syncMembers.size()) {
            // Iterate over syncMembers until a match if found
            // or we have exhausted the list
            found = syncMembers.get(i).split("keyxyz")[3].equals(memberNo);
            i++;
        }
        if (!found) {
            // Display only if not found
            System.out.println(memberNo);
        }
    }
}

比较arraylists的非相同元素

2 个答案: