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Question

我有两个清单

Collection<String> listOne = new ArrayList<String>();
Collection<String> listTwo = new ArrayList<String>();

for(PermissionDataObject permissionDataObject : selectedPermissions)
  listOne.add(permissionDataObject.getPermissionName());        

for(String selectedPermissions : selectedList)
  listTwo.add(selectedPermissions);

Collection<String> similarPermissions = new HashSet<String>(listOne);
Collection<String> differentPermissions = new HashSet<String>();

differentPermissions.addAll(listOne);
differentPermissions.addAll(listTwo);

similarPermissions.retainAll(listTwo);
differentPermissions.removeAll(similarPermissions);

System.out.printf("One:%s%nTwo:%s%nSimilar:%s%nDifferent:%s%n", listOne, listTwo, similarPermissions, differentPermissions);

输出为

One:[MENU_CREATION, ROLE_CREATION]
Two:[CREATE_BRANCH, CREATE_COUNTRY, MENU_CREATION, ROLE_CREATION]
Similar:[ROLE_CREATION, MENU_CREATION]
Different:[CREATE_BRANCH, CREATE_COUNTRY]

上面的代码完美运行，我得到了不同的集合

但我想检查 listTwo 和 differentPermissions，如果不同且 listTwo 等于 doSomething 如果不同且 listTwo 不等于 doSomthing

我有下面的代码，当不同的值有一个值时，它可以工作，如果不同的值有多个值，循环就会中断

for (Iterator<String> it = differentPermissions.iterator(); it.hasNext();){
    String diffString = it.next();
    String selectedString = selectedList.listIterator().next();

    log.info("---diffString---" + diffString);
    log.info("---selectedString---" + selectedString);

    if(diffString == selectedString){
         System.out.println("----permission added----" + diffString);
    }else{
         System.out.println("----permission deleted----" + diffString);
         }
    }

任何想法，对我来说都很棒......

谢谢普拉迪普

[编辑] 我的代码中的小变化请参阅下面的完整代码

Collection<String> listOne = Arrays.asList("MENU_CREATION", "ROLE_CREATION");
        Collection<String> listTwo = Arrays.asList("CREATE_BRANCH", "CREATE_COUNTRY", "MENU_CREATION", "ROLE_CREATION");

        Collection<String> similar = new HashSet<String>(listOne);
        Collection<String> different = new HashSet<String>();

        different.addAll(listOne);
        different.addAll(listTwo);

        similar.retainAll(listTwo);
        different.removeAll(similar);

         System.out.printf("One:%s%nTwo:%s%nSimilar:%s%nDifferent:%s%n", listOne, listTwo, similar, different);

        for (Iterator<String> it = different.iterator(); it.hasNext();) {

            String diffString = it.next();
            String selectedString = listTwo.iterator().next();

            if (diffString.equals(selectedString)) {
                System.out.println("----permission added----" + diffString);
            } else {
                System.out.println("----permission deleted----" + diffString);
            }
        }

Answer 1

快速查看代码后，尝试使用

diffString.equals(selectedString)

而不是

diffString == selectedString

。

那么，对于你的问题，我不太明白你要比较什么。您希望列表完美匹配，还是想检查列表中具有相同索引的项目是否匹配？

Answer 2

==

比较对象的引用而不是其内容。使用

diffString.equals(selectedString)

来比较

String

。

Answer 3

要找出集合中相同/不同的项目，我建议使用 Set 接口和标准集合操作，如并集、交集等。 您可以使用 google Guava 库来帮助您：http://code.google.com/p/guava-libraries/wiki/CollectionUtilitiesExplained

Set<String> wordsWithPrimeLength = ImmutableSet.of("one", "two", "three", "six", "seven", "eight");
Set<String> primes = ImmutableSet.of("two", "three", "five", "seven");

SetView<String> intersection = Sets.intersection(primes, wordsWithPrimeLength); // contains "two", "three", "seven"
// I can use intersection as a Set directly, but copying it can be more efficient if I use it a lot.
return intersection.immutableCopy();

您无需遍历您的收藏项目即可找到您需要的内容。希望对以后的实施有所帮助