Hibernate的UserType和一个定义的长度 [英] Hibernate UserType and a defined length

问题描述

  public class UUIDHibernateType implements UserType 
 {
 private static final int [] SQL_TYPES = new int [] {Types.CHAR}; 
 
 public int [] sqlTypes（）
 {
 return SQL_TYPES; 
} 
 
 // ... 
} 
  

我遇到的问题是，hibernate会生成一个CHAR类型为CHAR（1）的SQL脚本，但实际上我需要CHAR（36）。如何定义自定义类型的默认长度？

目前，我一直在定义像这样的sql类型：

 < id name =id
 type =org.openscada.ae.server.storage.jdbc.internal.UUIDHibernateType> 
< column name =IDlength =36sql-type =CHAR（36）not-null =true/> 
< generator class =assigned/> 
< / id> 
  

在这种情况下这不应该是个问题，但如果需要出现，我该怎么做如果有人有一个更好的主意如何透明地处理UUID和数据库agnostig，我很感激。

I have a hibernate Usertype something like this:

public class UUIDHibernateType implements UserType
{
    private static final int[] SQL_TYPES = new int[] { Types.CHAR };

    public int[] sqlTypes ()
    {
        return SQL_TYPES;
    }

    // ...
}

The problem I have is, that hibernate generates a sql script with the types CHAR(1) which is not correct, I would actually need CHAR(36). How would I define the default length of a custom type?

For the moment I'm stuck with defining the sql-type like this:

<id name="id"
    type="org.openscada.ae.server.storage.jdbc.internal.UUIDHibernateType">
    <column name="ID" length="36" sql-type="CHAR(36)" not-null="true" />
    <generator class="assigned" />
</id>

It shouldn't be a problem in this case, but how would I do it if the need arises?

PS: If someone has a better idea how to handle UUIDs transparently and database agnostig, I'm grateful.

解决方案

You can also map your UUID to a BigInteger (java.sql.Types.BIGINT) in your UserType if you don't care about how your UUID is represented in your database (in this case, it will be represented in base 10).

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