遍历迭代器返回相同的值

Question

我的循环没有中断并返回相同的值的问题

  public Map<String, String> getKeysByValue(Map<String, Map<String, Peple>> map, Collection<Peple> value) {

    Map<String, String> stringStringMap = new HashMap<>();

    int count = 0;

    while (value.iterator().hasNext()) {
        for (Map.Entry<String, Map<String, Peple>> entry : map.entrySet()) {
            for (Map.Entry<String, Peple> entry1 : entry.getValue().entrySet()) {
                String verified = value.iterator().next().gerVerfied();
                if (verified.equals("true")) { 
                    stringStringMap.put(entry1.getKey(), value.iterator().next().getName());

                }
            }
    }

这里的问题是，为每个键（value.iterator().next().getName()）放入映射中的相同值总是返回相同的字符串

Answer 1

此代码适用于以下输入：

      import java.util.*;

   public class Test {
   public static  void main(String[] ar){
    Map<String, People> entry1 = new HashMap<String , People>();
    People people1 = new People("true","name1");
    People people2 = new People("false","name2");
    People people3 = new People("true","name3");
    entry1.put("user1", people1);
    entry1.put("user2", people2);
    entry1.put("user3", people3);

    Map<String, People> entry2 = new HashMap<String , People>();
    People people4 = new People("true","name1");
    People people5 = new People("false","name2");
    People people6 = new People("true","name3");
    entry2.put("user1", people4);
    entry2.put("user2", people5);
    entry2.put("user3", people6);

    Map<String, Map<String, People>> map2 = new HashMap<String, Map<String, 
   People>>();
    map2.put("set1",entry1);
    map2.put("set2",entry2);

    Collection<People> strings = new ArrayList<>();
    strings.add(people1);
    strings.add(people6);
    strings.add(people5);
    strings.add(people3);
    strings.add(people2);

    getKeysByValue(map2, strings);

}

public static Map<String, String> getKeysByValue(Map<String, Map<String, People>> map, Collection<People> value) {

    Map<String, String> stringStringMap = new HashMap<>();

    int count = 0;
    Iterator<People> it = value.iterator();

    while (it.hasNext()) {
        People people = it.next();
        for (Map.Entry<String, Map<String, People>> entry : map.entrySet()) {
            for (Map.Entry<String, People> entry1 : entry.getValue().entrySet()) {

                String verified = people.getVerified();
                System.out.println(verified);
                System.out.println("Key : "+entry1.getKey() +" Value : "+entry1.getValue().getVerified());
                if (verified.equals(entry1.getValue().getVerified())) {
                   stringStringMap.put(entry1.getKey(), people.getName());
                }
            }
        }
    }
    System.out.println(stringStringMap);
    return stringStringMap;
   }
  }

People.java

  public class People {
String verified;

public  People(String verified, String name){
    this.verified = verified;
    this.name = name;
}

public String getVerified() {
    return verified;
}

public String getName() {
    return name;
}
String name;

}

Answer 2

我认为以下方法可以解决您的问题：

public Map<String, String> getKeysByValue(Map<String, Map<String, Peple>> map, Collection<Peple> value) {

            Map<String, String> stringStringMap = new HashMap<>();

            int count = 0;

            Iterator<People> peopleIterator = value.iterator();

            while (peopleIterator.hasNext()) {
                for (Map.Entry<String, Map<String, Peple>> entry : map.entrySet()) {
                    for (Map.Entry<String, Peple> entry1 : entry.getValue().entrySet()) {
                        People people = peopleIterator.next();
                        String verified = people.gerVerfied();
                        if (verified.equals("true")) {
                            stringStringMap.put(entry1.getKey(), people.getName());

                        }
                    }
                }

您需要将迭代器存储在一个变量中。

考虑下面的简单代码：

Collection<String> strings = new ArrayList<>();
        strings.add("value1");
        strings.add("value2");
        strings.add("value3");
        strings.add("value4");

        while (strings.iterator.hasNext())
        {
            System.out.println(strings.iterator.next());
        }

这将无限运行并且只会打印 value1，但如果你修改代码如下：

Collection<String> strings = new ArrayList<>();
strings.add("value1");
            strings.add("value2");
            strings.add("value3");
            strings.add("value4");

        Iterator<String> stringIterator = strings.iterator();

        while (stringIterator.hasNext())
        {
            System.out.println(stringIterator.next());
        }

运行流畅。

您可以在此处了解有关迭代器的更多信息：https://www.geeksforgeeks.org/how-to-use-iterator-in-java/

另外，我认为由于您在第二次调用时没有检查 hasNext() 两次调用迭代器，它可能会抛出 java.util.NoSuchElementException

Answer 3

每次调用value.iterator() 时，都会从头开始创建新的Iterator 对象，指向第一个元素。为了避免它，将调用它的第一个结果存储到局部变量中。

Iterator<Peple> it = value.iterator();

while(it.hasNext()){
  // your remaiing code
}

Answer 4

试试这个：

public Map<String, String> getKeysByValue(Map<String, Map<String, Peple>> map, Collection<Peple> value) {

    Map<String, String> stringStringMap = new HashMap<>();
    int count = 0;
    Iterator<Peple> iterator = value.iterator();
    while (iterator.hasNext()) {
        Peple p = iterator.next();
        for (Map.Entry<String, Map<String, Peple>> entry : map.entrySet()) {
            for (Map.Entry<String, Peple> entry1 : entry.getValue().entrySet()) {
                String verified = p.gerVerfied();
                if (verified.equals("true")) { 
                    stringStringMap.put(entry1.getKey(), p.getName());
                }
            }
        }
    }
}

Answer 5

您在该迭代器上调用了 next() 两次。它将向前移动并获取下一个人的名字，这可能无法验证。

如果它是经过验证的 Peple，我相信你只是想知道这个名字？

while (value.iterator().hasNext()) {
  for (Map.Entry<String, Map<String, Peple>> entry : map.entrySet()) {
    for (Map.Entry<String, Peple> entry1 : entry.getValue().entrySet()) {
      Peple nextPeple = value.iterator().next();
      String verified = nextPeple.gerVerfied();
      if (verified.equals("true")) { 
        stringStringMap.put(entry1.getKey(), nextPeple.getName());
      }
    }
  }
}