Java HashMap：如何为一个键添加多个值？

huangapple
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英文:

Java HashMap : How to add multiple values to one key?

问题

  1. 从dao获取了一个列表，我想把这个列表放入一个 `HashMap<Long, List<Object[]>>` 中，我的列表可以包含一个带有多个参数的服务，比如 `serviceId=3`。
  2. 在我的最终HashMap中，结果看起来像这样： `{1, [100], 2, [101], 3, [[102, B], [103, B], [104, C]]}`。我尝试了这段代码，但它没有起作用。
  4.     serviceId   paramId   type
  5.         1         100      A
  6.         2         101      A
  7.         3         102      B
  8.         3         103      B
  9.         3         104      C
  11. 代码：
  13.     List result = dao.getServiceParam();
  14.     HashMap<Long, List<Object[]>> mapArray = new HashMap<Long, List<Object[]>>();
  15.     List<Object[]> listObj = new ArrayList<Object[]>();
  16.     if (!result.isEmpty()) {
  17.         for (int i = 0; i < result.size(); i++) {
  18.             Object[] line = (Object[]) result.get(i);
  19.             if ((BigDecimal) line[0] != null) {
  20.                 listObj.add(line);
  21.                 mapArray.put(new Long(((BigDecimal) line[0]).longValue()), listObj);
  22.             }
  23.         }
  24.     }
英文:

I have a list from dao, I want to put this list in a HashMap&lt;Long,List&lt;Object[]&gt;&gt;, my list can contain a service which have multiple parameters like the serviceId=3.
In my final HashMap, the result looks like : {1,[100],2[101],3=[[102,B],[103,B],[104,C]]}. I tried with this code but it didn't work.

  1. serviceId   paramId   type
  2.     1         100      A
  3.     2         101      A
  4.     3         102      B
  5.     3         103      B
  6.     3         104      C

Code:

  1. List result = dao.getServiceParam();
  2. HashMap&lt;Long,List&lt;Object[]&gt;&gt; mapArray = new HashMap&lt;Long, List&lt;Object[]&gt;&gt;();
  3. List&lt;Object[]&gt; listObj = new ArrayList&lt;Object[]&gt;();
  4. if(!result.isEmpty()) {				
  5. 	for (int i=0; i&lt; result.size(); i++) {
  6. 	    Object[] line = (Object[])result.get(i);		
  7.         if ((BigDecimal) line[0]!=null) {
  8. 	        istObj.add(line);
  9. 	        mapArray .put(new Long(((BigDecimal) line[0]).longValue()), listObj);	
  10.         }
  11.     }
  12. }

答案1

得分: 0

代替使用对象数组，创建您自己的POJO来表示数据，然后将其存储在列表中。然后，我们可以使用Collectors＃grouping按serviceId进行分组，然后Collect到一个列表中的POJO对象列表。

  1. class POJO {
  2.     private final long serviceId;
  4.     private final int paramId;
  6.     private final String type;
  8.     // 构造函数，获取器
  9. }
  1. Map<Long, List<POJO>> map = new HashMap<>();
  1. Map<Long, List<POJO>> mapArray = result.stream()
  2.         .collect(Collectors.groupingBy(POJO::getServiceId, Collectors.toList()));
英文:

Instead of using an array of Object, create your own POJO to represent the data then store it in a list. We can then take the List of POJO objects and create a map using Collectors#grouping to group by the serviceId and then Collect to a List.

  1. class POJO {
  2.     private final long serviceId;
  4.     private final int paramId;
  6.     private final String type;
  8.     // constructor, getters
  9. }
  1. Map&lt;Long, List&lt;POJO&gt;&gt; map = new HashMap&lt;&gt;();
  1.         Map&lt;Long, List&lt;POJO&gt;&gt; mapArray = result.stream()
  2.                 .collect(Collectors.groupingBy(POJO::getServiceId, Collectors.toList()));

答案2

得分: 0

不要重复使用相同的 ArrayList 引用。创建 new ArrayList&lt;Object[]&gt;();，并将 line 添加到其中。操作如下：

  1. List result = dao.getServiceParam();
  2. HashMap&lt;Long,List&lt;Object[]&gt;&gt; mapArray = new HashMap&lt;Long, List&lt;Object[]&gt;&gt;();
  3. List&lt;Object[]&gt; listObj = null;
  4. Object[] line = null, nextLine = null;
  5. Long id = Long.valueOf(0), nextId = Long.valueOf(0);
  6. if(!result.isEmpty()) {             
  7.     for (int i=0; i&lt; result.size(); i++) {
  8.     	listObj = new ArrayList&lt;Object[]&gt;();
  9.     	do{
  10. 	        line = (Object[])result.get(i);        
  11. 	        if ((BigDecimal) line[0]!=null) {	            
  12. 	            listObj.add(line);	
  13. 	            id = new Long(((BigDecimal) line[0]).longValue());              
  14. 	        }
  15. 	        i++;
  16. 	        if(i&lt;result.size()){
  17. 		        nextLine = (Object[])result.get(i);        
  18. 		        if ((BigDecimal) nextLine[0]!=null) {	            
  19. 		            nextId = new Long(((BigDecimal) nextLine[0]).longValue());              
  20. 		        }
  21. 	        } 
  22. 	    }while(id.equals(nextId));    
  23. 	    mapArray.put(id, listObj);
  24. 	    i--;
  25.     }
  26. }

如有疑问或问题，请随时进行评论。

英文:

Do not reuse the same reference of ArrayList. Create new ArrayList&lt;Object[]&gt;(); and add line to it. Do it as follows:

  1. List result = dao.getServiceParam();
  2. HashMap&lt;Long,List&lt;Object[]&gt;&gt; mapArray = new HashMap&lt;Long, List&lt;Object[]&gt;&gt;();
  3. List&lt;Object[]&gt; listObj = null;
  4. Object[] line = null, nextLine = null;
  5. Long id = Long.valueOf(0), nextId = Long.valueOf(0);
  6. if(!result.isEmpty()) {             
  7.     for (int i=0; i&lt; result.size(); i++) {
  8.     	listObj = new ArrayList&lt;Object[]&gt;();
  9.     	do{
  10. 	        line = (Object[])result.get(i);        
  11. 	        if ((BigDecimal) line[0]!=null) {	            
  12. 	            listObj.add(line);	
  13. 	            id = new Long(((BigDecimal) line[0]).longValue());              
  14. 	        }
  15. 	        i++;
  16. 	        if(i&lt;result.size()){
  17. 		        nextLine = (Object[])result.get(i);        
  18. 		        if ((BigDecimal) nextLine[0]!=null) {	            
  19. 		            nextId = new Long(((BigDecimal) nextLine[0]).longValue());              
  20. 		        }
  21. 	        } 
  22. 	    }while(id.equals(nextId));    
  23. 	    mapArray.put(id, listObj);
  24. 	    i--;
  25.     }
  26. }

Feel free to comment in case of any doubt/issue.

答案3

得分: 0

  1. if (!result.isEmpty()) {
  2.     for (int i = 0; i < result.size(); i++) {
  3.         Object[] line = (Object[]) result.get(i);
  4.         if (((BigDecimal) line[0]) != null) {
  5.             Long longVal = new Long(((BigDecimal) line[0]).longValue());
  6.             if (mapArray.containsKey(longVal)) {
  7.                 List<Object[]> tempList = mapArray.get(longVal);
  8.                 tempList.add(line);
  9.                 mapArray.put(longVal, tempList);
  10.             } else {
  11.                 List<Object[]> tempList = new ArrayList<>();
  12.                 tempList.add(line);
  13.                 mapArray.put(longVal, tempList);
  14.             }
  15.         }
  16.     }
  17. }
英文: 
  1. if(!result.isEmpty()) {             
  2.  for (int i=0; i&lt; result.size(); i++) {
  3.   Object[] line=(Object[])result.get(i);     
  4.       if((BigDecimal) line[0]!=null) {
  5.       Long longVal = new Long(((BigDecimal) line[0]).longValue());
  6.        if(mapArray.containsKey(longVal){
  7.          List&lt;Object[]&gt; tempList = mapArray.get(longVal);
  8.          tempList.add(line);
  9.          mapArray.put(longVal, tempList); 
  10.       }
  11.        else{
  12.          List&lt;Object[]&gt; tempList = new ArrayList&lt;&gt;();
  13.          tempList.add(line);
  14.          mapArray.put(longVal,tempList);
  15.       }
  16.     }
  17.   }
  18. }
  19.

huangapple
  • 本文由 发表于 2020年4月6日 20:16:29
  • 转载请务必保留本文链接：https://go.coder-hub.com/61059660.html
  • java
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