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试图理解我从对象相等中得到的结果......好吧，首先让我描述需要从数据库中的两个两个表中过滤重复对象的问题（Hibenate所有的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;  
    }  
}

所以在我的服务类中，我在 2 个表上调用 Query 并创建对象列表迭代它

在其中一个列表中创建一个嵌套迭代和第二个列表

所以在我的服务类中，我在 2 个表上调用 Query 并创建自定义列表迭代列表和对象

所以在我的服务类中，我在 2 个表上调用 Query 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());

所以这是我不理解的部分，我在对象上实现了相等方法，但相等返回一两个匹配项... 2012-09-01 18:49:09,910 INFO [com.equality.service.CopareToServices] - （线程 1008调用 CopareToServices.compareObjects) Equality 方法只找到一个重复记录 >>>

但是当最后我在列表上实现 removeAll 时，结果

这是在 heshSet 2012-09-01 18:49:09,918 INFO [com.equality.service.CopareToServices] 上实施 removeAll 之后 - （线程 1008 调用 CopareToServices.compareObjects）

删除了 171 个听起来正确的重复项...

如何在列表上实现 removeAll 正确过滤列表但相等方法不起作用

尝试使用 apache commons 的 equals builder 也具有相同的输出

真诚地感谢我能得到的任何帮助

4

1 回答 1

1

如果我理解得很好，我认为问题是：

for (AsignoreTypes a1: res1) {
    f = ass.equals(a1);
}

因为这就像只测试res1.

编辑：更正建议：

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());
}

参见List#contains(Object)

于 2012-09-02T05:45:31.687 回答