Question

我有一个像这样的String数组：

String[] a={"Hi","Hello","Dear","Jungle"};

我有一个像这样的HashMap：

Map<String, String> map=new HashMap<>();
map.put("Hi", "1");
map.put("Hello", "2");
map.put("Dear", "3");
map.put("Jungle", "4");
map.put("Sai", "5");
map.put("Surya", "6");

HashMap是我比较的基础。如果我们看到，String数组错过了两个字符串 SAI 和 SURYA 。

我正在遍历hashmap并使用String数组检查，如下所示：

int c=0;
for(String key: map.keySet()){
    c=0;
    for (int i = 0; i < a.length; i++) {

        if(key.equals(a[i])){
            c++;
            System.out.println("App :"+ key+"present in DB , SO it is passed.");
        }               
    }
    if(c==0){
        System.out.println("App not present in DB, SO it might have failed");
    }
}

这是我的完整计划：

import java.util.HashMap;
import java.util.Map;


public class TestingApplication {

    public static void main(String[] args) {

        String[] a={"Hi","Hello","Dear","Jungle"};
        String[] status={"Passed","Passed","Passed","Passed"};

        Map<String, String> map=new HashMap<>();
        map.put("Hi", "1");
        map.put("Hello", "2");
        map.put("Dear", "3");
        map.put("Jungle", "4");
        map.put("Sai", "5");
        map.put("Surya", "6");
        int c=0;
        for(String key: map.keySet()){
            c=0;
            for (int i = 0; i < a.length; i++) {

                if(key.equals(a[i])){
                    c++;
                    System.out.println("App:"+ a[i]+" present in DB , SO it is passed.");
                }
            }
            if(c==0){
                System.out.println("App not present in DB, SO it might have failed");
            }
        }
    }
}

我的输出是：

App not present in DB, SO it might have failed
App:Hi present in DB , SO it is passed.
App:Dear present in DB , SO it is passed.
App:Hello present in DB , SO it is passed.
App:Jungle present in DB , SO it is passed.
App not present in DB, SO it might have failed

1.。）我们如何知道数组中缺少哪个KEY？ 2.）我们如何将缺少的String添加到数组中？

编辑：我的问题不是迭代HashMap。我的问题是如何找出差异并将它们存储回数据结构中。如果您愿意，请阅读我的问题，然后将其标记为重复。

Answer 1

使用Guava库实现这一目标非常简单，请参阅文档以获取更多信息：https://github.com/google/guava/wiki/CollectionUtilitiesExplained

使用Java-8功能，我打印它但你可以对itemsNotFoundInArray进行迭代并添加到缺少项目的数组。

final Set<String> itemsFromArray = Sets.newHashSet(a);
final Set<String> itemsFromMap= map.keySet();
Sets.SetView<String> itemsNotInArray = Sets.difference(itemsFromMap, itemsFromArray);
itemsNotInArray.forEach(System.out::println);

Answer 2

首先，如果你想在数组中添加一些内容，那么除非你事先创建一个大尺寸的数组，否则你需要动态数组。 ArrayList将是一个很好的候选人。然后，一旦你创建了一个ArrayList，那么实现你的任务将是2或3个班轮代码。

    String[] staticArray = new String[] { "Hi","Hello","Dear","Jungle"};
    ArrayList myList = Arrays.asList(staticArray);
    for(String key: map.keySet()){
    if(!myList.contains(key){
         myList.add(key);    
      }
    }

Answer 3

您只需在代码中更改一些内容，请尝试以下代码

 public class TestingApplication 
    {
        public static void main(String[] args) 
        {
            String[] a = { "Hi", "Hello", "Dear", "Jungle" };
            Map<String, String> map = new HashMap<>();
            map.put("Hi", "1");
            map.put("Hello", "2");
            map.put("Dear", "3");
            map.put("Jungle", "4");
            map.put("Sai", "5");
            map.put("Surya", "6");
            List<String> missedKeyList = new ArrayList<>();
            for (String key : map.keySet()) 
            {
                int c = 0;
                for (int i = 0; i < a.length; i++) 
                {
                    if (key.equals(a[i])) 
                    {
                        c++;
                        System.out.println("App: " + a[i] + " present in DB , So it is passed.");
                        break;
                    }
                }

                if (c == 0) 
                {
                    System.out.println("App: " + key + " not present in DB, So it might have failed");
                    missedKeyList.add(key);
                }
            }

            String[] allKey = Arrays.copyOfRange(a, 0, a.length + missedKeyList.size());
            int counter = a.length;
            for(String missedKey : missedKeyList)
            {
                allKey[counter] = missedKey;
                counter++;
            }
        }
    }

比较字符串数组和HashMap时如何找到差异？

3 个答案: