双向一对一映射@OneToOne

双向一对一的实例我已经上传到GitHub，entrance项目上了，感兴趣的可以下载下来跑跑，这里讲两个在运行过程中遇到的问题。

问题一：上一篇博客的最后我留下了问题。一对一关联查询注解@OneToOne的实例详解

问题点：文章的最后，单向一对一程序在执行save语句的时候，分别执行了下面的语句：

Hibernate: select person0_.id as id1_2_1_, person0_.name as name2_2_1_, person0_.pet_id as pet_id3_2_1_, pet1_.id as id1_3_0_, pet1_.pet_class as pet_clas2_3_0_, pet1_.pet_name as pet_name3_3_0_ from person person0_ inner join pet pet1_ on person0_.pet_id=pet1_.id where person0_.id=?
Hibernate: insert into pet (pet_class, pet_name) values (?, ?)
Hibernate: insert into person (name, pet_id) values (?, ?)
Hibernate: insert into pet (pet_class, pet_name) values (?, ?)
Hibernate: update person set name=?, pet_id=? where id=?

这里我感觉很奇怪的地方就是执行了内连接，所以导致都没有查到。

再看看双向一对一执行的语句：

Hibernate: select husband0_.hid as hid1_1_1_, husband0_.hid_card as hid_card2_1_1_, husband0_.hname as hname3_1_1_, husband0_.w_id as w_id4_1_1_, wife1_.wid as wid1_6_0_, wife1_.wid_card as wid_card2_6_0_, wife1_.wname as wname3_6_0_ from husband husband0_ left outer join wife wife1_ on husband0_.w_id=wife1_.wid where husband0_.hid=?
Hibernate: insert into wife (wid_card, wname) values (?, ?)
Hibernate: insert into wife (wid_card, wname) values (?, ?)
Hibernate: update husband set hid_card=?, hname=?, w_id=? where hid=?
Hibernate: update husband set hid_card=?, hname=?, w_id=? where hid=?

双向一对一连接的话第一句执行的是左连接，所以能查到数据，这样就能直接更新了。

问题二：异常：在使用fastJson作为工具，将数据库查到的数据转换成fastJson输出，抛出了：java.lang.StackOverflowError: null

这个异常就是堆栈溢出，可以很明显的看出来，是因为husband对象与wife之前嵌套，在序列化的时候出现了这个异常。网上搜了一些，大多是List集合的嵌套，从中找到些蛛丝马迹。解决的方法：

方法一：把wife中husband的属性不序列化，加上属性transient,加上序列属性serialVersionUID

	private static final long serialVersionUID = 8091602122698339709L;
	@Id
	// id自动生成
	@GeneratedValue
	@Column(name = "wid")
	private Long wid;
	@Column(name = "wname")
	private String wname;
	@Column(name = "wid_card")
	private String widCard;
	
	 @OneToOne(mappedBy = "wife",cascade=CascadeType.ALL)
	 @JsonBackReference
	 private transient Husband husband;

方法二：重新字节写toString方法，不要使用系统自己生成的（这里也有技巧）。

husband.toString()

@Override
	public String toString() {
		return "{\"hid\":\"" + hid + "\",\"hname\":\"" + hname
				+ "\",\"hidCard\":\"" + hidCard + "\",\"wife\":" + wife
				+ "} ";
	}

wife.toString()

@Override
	public String toString() {
		return "{\"wid\":\"" + wid + "\",\"wname\":\"" + wname
				+ "\",\"widCard\":\"" + widCard + "\"} ";
	}
	

我的代码中采用了第二种方法。