试图理解我从对象相等中得到的结果...... 好的，首先让我描述一下问题 需要从数据库中过滤两个两个表中的重复对象(Hibenate所有cluses都已设置)预设条件是firstName，lastName，dateOfbirth
package com.equality.types;
import com.google.common.base.Objects;
import java.util.Date;
public class AsignoreTypes{
private String firstname;
private String lastname;
private Date doa;
public String getFirstname() {
return firstname;
}
public void setFirstname(String firstname) {
this.firstname = firstname;
}
public String getLastname() {
return lastname;
}
public void setLastname(String lastname) {
this.lastname = lastname;
}
public Date getDoa() {
return doa;
}
public void setDoa(Date doa) {
this.doa = doa;
}
/* @Override
public boolean equals(Object obj) {
if (obj instanceof AsignoreTypes == false) {
return false;
}
if (this == obj) {
return true;
}
AsignoreTypes other = (AsignoreTypes) obj;
return new EqualsBuilder().append(this.firstname , other.firstname)
.append(this.lastname, other.lastname)
.append(this.doa , other.doa).isEquals();
}
@Override
public int hashCode() {
return new HashCodeBuilder().append(this.firstname)
.append(this.lastname)
.append(this.doa)
.hashCode();
}
*/
@Override
public int hashCode(){
return Objects.hashCode(firstname, lastname, doa);
}
@Override
public boolean equals(final Object obj){
if(obj instanceof AsignoreTypes){
final AsignoreTypes other = (AsignoreTypes) obj;
return Objects.equal(firstname, other.firstname)
&&Objects.equal(lastname, other.lastname)
&& Objects.equal(doa, other.doa);
} else{
return false;
}
}
因此，在我的服务类中，我对 2 个表调用 Query 并创建对象列表，对其进行迭代
在其中一个列表中创建第二个列表的嵌套迭代
因此，在我的服务类中，我对 2 个表调用查询并创建自定义列表迭代列表和对象
因此，在我的服务类中，我对 2 个表调用查询并创建自定义列表迭代列表和对象
Session session = service.getDataServiceManager().getSession();
SimpleDateFormat formatter = new SimpleDateFormat("yyyy-MM-dd");
Query query = session.createQuery("select upper(ass.firstName), upper(ass.lastName), ass.doa from AssignorInfo ass");
List < Object > ass1 = null;
ass1 = query.list();
List < AsignoreTypes > res1 = null;
res1 = new ArrayList < AsignoreTypes > ();
Iterator < Object > it = ass1.iterator();
while (it.hasNext()) {
Object[] row = (Object[]) it.next();
AsignoreTypes ass = new AsignoreTypes();
if (row[0] != null) ass.setFirstname(row[0].toString());
if (row[1] != null) ass.setLastname(row[1].toString());
if (row[2] != null) ass.setDoa(formatter.parse(row[2].toString()));
res1.add(ass);
}
log(INFO, "List one size " + res1.size());
Query q = session.createQuery("select upper(g.fname), upper(g.lname), g.doa from GeneralInfo g where g.arbitrator like '%Bulel%' ");
List < Object > ass2 = null;
ass2 = q.list();
log(INFO, " Size from wizard " + ass2.size());
List < AsignoreTypes > res2 = null;
res2 = new ArrayList < AsignoreTypes > ();
Iterator < Object > iterator = ass2.iterator();
while (iterator.hasNext()) {
Boolean f = null;
Object[] row = (Object[]) iterator.next();
AsignoreTypes ass = new AsignoreTypes();
if (row[0] != null) ass.setFirstname(row[0].toString());
if (row[1] != null) ass.setLastname(row[1].toString());
if (row[2] != null) ass.setDoa(formatter.parse(row[2].toString()));
for (AsignoreTypes a1: res1) {
f = ass.equals(a1);
}
if (f == false) {
res2.add(ass);
log(INFO, "Got matchig element " + ass.getLastname());
}
}
log(INFO, "List tow size " + res2.size() + " List hash set size ");
HashSet < AsignoreTypes > asigors = new HashSet < AsignoreTypes > ();
HashSet < AsignoreTypes > wiz = new HashSet < AsignoreTypes > ();
asigors.addAll(res1);
wiz.addAll(res2);
wiz.removeAll(asigors);
log(INFO, "Added to hash set " + asigors.size());
log(INFO, "Added assignors to the list " + wiz.size());
所以这是我不明白的部分，我在对象上实现了相等方法，但相等返回一两个匹配... 2012-09-01 18:49:09,910 INFO [com.equality.service.CopareToServices] - (线程 1008 调用 CopareToServices.compareObjects) 相等方法只找到一条重复记录>>>
但是当最后我在列表上实现removeAll时，结果
这是在heshSet上实现removeAll之后的结果 2012-09-01 18:49:09,918 INFO [com.equality.service.CopareToServices] - (线程 1008 调用 CopareToServices.compareObjects)
171 个重复项被删除，听起来不错......
为什么在列表上实现removeAll可以正确过滤列表，但相等方法不起作用
也尝试了使用 apache commons 的 equals 构建器，具有相同的输出
我能得到任何帮助吗？ 真诚的
最佳答案
如果我理解得好的话，我认为问题是:
for (AsignoreTypes a1: res1) {
f = ass.equals(a1);
}
因为这就像只测试 res1 的最后一个元素。
编辑:修正建议:
private List<AsignoreTypes> getAssignoreTypesList(final String sql) {
final List<AsignoreTypes> result = new ArrayList<AsignoreTypes>();
final SimpleDateFormat formatter = new SimpleDateFormat("yyyy-MM-dd");
final Session session = service.getDataServiceManager().getSession();
final Query query = session.createQuery(sql);
final List<?> rows = query.list();
for (final Object row : rows) {
final Object[] columns = (Object[]) row;
final AsignoreTypes ass = new AsignoreTypes();
ass.setFirstname((String) columns[0]);
ass.setLastname((String) columns[1]);
if (columns[2] != null) {
try {
ass.setDoa(formatter.parse(columns[2].toString()));
} catch (final ParseException e) {
// ignore?
}
}
result.add(ass);
}
return result;
}
public void foo()
{
final List<AsignoreTypes> assignorInfos = getAsignoreTypesList("select upper(ass.firstName), upper(ass.lastName), ass.doa from AssignorInfo ass");
log(INFO, "List one size " + assignorInfos.size());
final List<AsignoreTypes> generalInfos = getAsignoreTypesList("select upper(g.fname), upper(g.lname), g.doa from GeneralInfo g where g.arbitrator like '%Bulel%' ");
log(INFO, "Size from wizard " + generalInfos.size());
final List<AsignoreTypes> filtered = new ArrayList<AsignoreTypes>();
for (final AsignoreTypes asignoreTypes : generalInfos) {
if (!assignorInfos.contains(asignoreTypes)) {
filtered.add(asignoreTypes);
log(INFO, "Got matching element " + asignoreTypes.getLastname());
}
}
log(INFO, "Filtered : " + filtered.size() + "/" + generalInfos.size());
}
关于java - 对象平等 Google guava 和 Apache Commons，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/12233570/