英文:
Java HashMap : How to add multiple values to one key?
问题
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`。
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。
serviceId paramId type
1 100 A
2 101 A
3 102 B
3 103 B
3 104 C
代码：
List result = dao.getServiceParam();
HashMap<Long, List<Object[]>> mapArray = new HashMap<Long, List<Object[]>>();
List<Object[]> listObj = new ArrayList<Object[]>();
if (!result.isEmpty()) {
for (int i = 0; i < result.size(); i++) {
Object[] line = (Object[]) result.get(i);
if ((BigDecimal) line[0] != null) {
listObj.add(line);
mapArray.put(new Long(((BigDecimal) line[0]).longValue()), listObj);
}
}
}
英文:
I have a list from dao, I want to put this list in a HashMap<Long,List<Object[]>>, 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.
serviceId paramId type
1 100 A
2 101 A
3 102 B
3 103 B
3 104 C
Code:
List result = dao.getServiceParam();
HashMap<Long,List<Object[]>> mapArray = new HashMap<Long, List<Object[]>>();
List<Object[]> listObj = new ArrayList<Object[]>();
if(!result.isEmpty()) {
for (int i=0; i< result.size(); i++) {
Object[] line = (Object[])result.get(i);
if ((BigDecimal) line[0]!=null) {
istObj.add(line);
mapArray .put(new Long(((BigDecimal) line[0]).longValue()), listObj);
}
}
}
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。
class POJO {
private final long serviceId;
private final int paramId;
private final String type;
// 构造函数，获取器
}
Map<Long, List<POJO>> map = new HashMap<>();
Map<Long, List<POJO>> mapArray = result.stream()
.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.
class POJO {
private final long serviceId;
private final int paramId;
private final String type;
// constructor, getters
}
Map<Long, List<POJO>> map = new HashMap<>();
Map<Long, List<POJO>> mapArray = result.stream()
.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<Object[]>();，并将 line 添加到其中。操作如下：
List result = dao.getServiceParam();
HashMap<Long,List<Object[]>> mapArray = new HashMap<Long, List<Object[]>>();
List<Object[]> listObj = null;
Object[] line = null, nextLine = null;
Long id = Long.valueOf(0), nextId = Long.valueOf(0);
if(!result.isEmpty()) {
for (int i=0; i< result.size(); i++) {
listObj = new ArrayList<Object[]>();
do{
line = (Object[])result.get(i);
if ((BigDecimal) line[0]!=null) {
listObj.add(line);
id = new Long(((BigDecimal) line[0]).longValue());
}
i++;
if(i<result.size()){
nextLine = (Object[])result.get(i);
if ((BigDecimal) nextLine[0]!=null) {
nextId = new Long(((BigDecimal) nextLine[0]).longValue());
}
}
}while(id.equals(nextId));
mapArray.put(id, listObj);
i--;
}
}
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<Object[]>(); and add line to it. Do it as follows:
List result = dao.getServiceParam();
HashMap<Long,List<Object[]>> mapArray = new HashMap<Long, List<Object[]>>();
List<Object[]> listObj = null;
Object[] line = null, nextLine = null;
Long id = Long.valueOf(0), nextId = Long.valueOf(0);
if(!result.isEmpty()) {
for (int i=0; i< result.size(); i++) {
listObj = new ArrayList<Object[]>();
do{
line = (Object[])result.get(i);
if ((BigDecimal) line[0]!=null) {
listObj.add(line);
id = new Long(((BigDecimal) line[0]).longValue());
}
i++;
if(i<result.size()){
nextLine = (Object[])result.get(i);
if ((BigDecimal) nextLine[0]!=null) {
nextId = new Long(((BigDecimal) nextLine[0]).longValue());
}
}
}while(id.equals(nextId));
mapArray.put(id, listObj);
i--;
}
}
Feel free to comment in case of any doubt/issue.
答案3
得分: 0
if (!result.isEmpty()) {
for (int i = 0; i < result.size(); i++) {
Object[] line = (Object[]) result.get(i);
if (((BigDecimal) line[0]) != null) {
Long longVal = new Long(((BigDecimal) line[0]).longValue());
if (mapArray.containsKey(longVal)) {
List<Object[]> tempList = mapArray.get(longVal);
tempList.add(line);
mapArray.put(longVal, tempList);
} else {
List<Object[]> tempList = new ArrayList<>();
tempList.add(line);
mapArray.put(longVal, tempList);
}
}
}
}
英文:
if(!result.isEmpty()) {
for (int i=0; i< result.size(); i++) {
Object[] line=(Object[])result.get(i);
if((BigDecimal) line[0]!=null) {
Long longVal = new Long(((BigDecimal) line[0]).longValue());
if(mapArray.containsKey(longVal){
List<Object[]> tempList = mapArray.get(longVal);
tempList.add(line);
mapArray.put(longVal, tempList);
}
else{
List<Object[]> tempList = new ArrayList<>();
tempList.add(line);
mapArray.put(longVal,tempList);
}
}
}
}
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