先决条件:Java 7 和 SQL Server 2016。
我正在从 SQL Server 获取存储过程返回的结果集
结果集:
RId ConId ActNbr StageId Qty HoursInStage HoursPassed HourlyQty FlowedQty
--- ------ ------ ------- ------ ------------ ----------- --------- ---------
50 6814 77 1 24000 24 0 NULL NULL
50 6814 77 2 36000 19 5 NULL NULL
50 6814 77 3 48000 15 9 NULL NULL
50 6814 77 4 60000 11 13 NULL NULL
50 6814 77 6 60000 24 0 NULL NULL
50 6855 33 1 0 24 0 NULL NULL
50 6855 33 2 15000 19 5 NULL NULL
50 6855 33 3 15000 15 9 NULL NULL
50 6855 33 4 15000 11 13 NULL NULL
50 6855 33 6 20000 24 0 NULL NULL
50 176892 10 1 0 24 0 NULL NULL
50 176892 10 2 0 19 5 NULL NULL
50 176892 10 3 0 15 9 NULL NULL
50 176892 10 4 0 11 13 NULL NULL
50 176892 10 6 0 24 0 NULL NULL
50 176892 47 1 0 24 0 NULL NULL
50 176892 47 2 0 19 5 NULL NULL
50 176892 47 3 0 15 9 NULL NULL
50 176892 47 4 0 11 13 NULL NULL
50 176892 47 6 0 24 0 NULL NULL
我创建了一个 POJO 类来保存结果集。
我想创建一个 Hashmap 对象，其中 Key 作为 RId、ConId、ActNbr 和 Value 作为 List，其中包含给定键的所有值。
那么，我可以使用 HashMap 来做到这一点吗？我想我可以使用将其他 map 保存为值的 map 。但不确定如何将公共(public)键值项制作为列表。
Java 类:HoursQty
public class BurnProfileHourlyNomData {
private Integer stageId;
private Integer qty;
private Integer hoursInStage;
private Integer hoursPassed;
private Integer hourlyQty;
private Integer FlowedQty;
}
逻辑:
Map<Integer, Map<Integer, Map<Integer, HoursQty>>> hourlyInfo = new HashMap<>();
for(DatabaseRecord row : rows) {
Integer rId = (Integer)row.get("RId");
Integer conId = (Integer)row.get("ConId");
Integer actNbr = (Integer)row.get("ActNbr");
Integer stageId = (Integer)row.get("StageId");
Integer qty = (Integer)row.get("Qty");
Integer hoursInStage = (Integer)row.get("HoursInStage");
Integer hoursPassed = (Integer)row.get("HoursPassed");
Integer hourlyQty = (Integer)row.get("HourlyQty");
Integer flowedQty = (Integer)row.get("FlowedQty");
HoursQty impl = new HoursQty();
impl.setRId(rId);
/* Performed all set methods to assign values */
.....
Map<Integer, HoursQty> actNbrMap = new HashMap<Integer, HoursQty>();
actNbrMap.put(actNbr, impl);
Map<Integer, Map<Integer, HoursQty>> conIdMap = new HashMap<Integer, Map<Integer, HoursQty>>();
conIdMap.put(conId, actNbrMap);
hourlyInfo.put(rId, conIdMap);
}
尝试使用 entryMap 键和值读取此内容，但得到的是单独的行值。
不确定如何根据 rId、conId 和 actNbr 作为键对所有列表进行分组，并检索与列表具有相同键的所有行。
最佳答案
你必须像这样设计你的类。
类 AccountKey 将用作 HashMap 中的键。代码如下。
public class AccountKey {
private Integer rId;
private Integer conId;
private Integer actNbr;
public Integer getrId() {
return rId;
}
public void setrId(Integer rId) {
this.rId = rId;
}
public Integer getConId() {
return conId;
}
public void setConId(Integer conId) {
this.conId = conId;
}
public Integer getActNbr() {
return actNbr;
}
public void setActNbr(Integer actNbr) {
this.actNbr = actNbr;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
AccountKey that = (AccountKey) o;
if (rId != null ? !rId.equals(that.rId) : that.rId != null) return false;
if (conId != null ? !conId.equals(that.conId) : that.conId != null) return false;
return actNbr != null ? actNbr.equals(that.actNbr) : that.actNbr == null;
}
@Override
public int hashCode() {
int result = rId != null ? rId.hashCode() : 0;
result = 31 * result + (conId != null ? conId.hashCode() : 0);
result = 31 * result + (actNbr != null ? actNbr.hashCode() : 0);
return result;
}
}
BurnProfileHourlyNomData 类将用作 HashMap 中的值。
public class BurnProfileHourlyNomData {
private Integer stageId;
private Integer qty;
private Integer hoursInStage;
private Integer hoursPassed;
private Integer hourlyQty;
private Integer FlowedQty;
public Integer getStageId() {
return stageId;
}
public void setStageId(Integer stageId) {
this.stageId = stageId;
}
public Integer getQty() {
return qty;
}
public void setQty(Integer qty) {
this.qty = qty;
}
public Integer getHoursInStage() {
return hoursInStage;
}
public void setHoursInStage(Integer hoursInStage) {
this.hoursInStage = hoursInStage;
}
public Integer getHoursPassed() {
return hoursPassed;
}
public void setHoursPassed(Integer hoursPassed) {
this.hoursPassed = hoursPassed;
}
public Integer getHourlyQty() {
return hourlyQty;
}
public void setHourlyQty(Integer hourlyQty) {
this.hourlyQty = hourlyQty;
}
public Integer getFlowedQty() {
return FlowedQty;
}
public void setFlowedQty(Integer flowedQty) {
FlowedQty = flowedQty;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
BurnProfileHourlyNomData that = (BurnProfileHourlyNomData) o;
if (stageId != null ? !stageId.equals(that.stageId) : that.stageId != null) return false;
if (qty != null ? !qty.equals(that.qty) : that.qty != null) return false;
if (hoursInStage != null ? !hoursInStage.equals(that.hoursInStage) : that.hoursInStage != null)
return false;
if (hoursPassed != null ? !hoursPassed.equals(that.hoursPassed) : that.hoursPassed != null)
return false;
if (hourlyQty != null ? !hourlyQty.equals(that.hourlyQty) : that.hourlyQty != null)
return false;
return FlowedQty != null ? FlowedQty.equals(that.FlowedQty) : that.FlowedQty == null;
}
@Override
public int hashCode() {
int result = stageId != null ? stageId.hashCode() : 0;
result = 31 * result + (qty != null ? qty.hashCode() : 0);
result = 31 * result + (hoursInStage != null ? hoursInStage.hashCode() : 0);
result = 31 * result + (hoursPassed != null ? hoursPassed.hashCode() : 0);
result = 31 * result + (hourlyQty != null ? hourlyQty.hashCode() : 0);
result = 31 * result + (FlowedQty != null ? FlowedQty.hashCode() : 0);
return result;
}
}
HashMap的结构会是这样的。
Map<AccountKey,BurnProfileHourlyNomData> actBurnDataMap = new HashMap<>();
从 stroed 过程接收到数据后，填充两个对象并使用上面给出的 Map。
关于Java:我们可以使用多HashMap来对结果集进行分组吗，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/57399051/