java Hibernate - 如何提供到整数类型的正确映射？
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Hibernate - How to provide right mapping to integer type?
提问by Comic Sans MS Lover
I'm executing my maven build and it throws this exception:
我正在执行我的 Maven 构建并抛出此异常：
Last cause: Wrong column type in x.clients for column type. Found: tinyint, expected: integer
Last cause: Wrong column type in x.clients for column type. Found: tinyint, expected: integer
I'm mapping like this:
我是这样映射的：
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;
And I'm creating the column using InnoDB like this: id int NOT NULL UNIQUE AUTO_INCREMENT
我正在使用 InnoDB 创建列，如下所示： id int NOT NULL UNIQUE AUTO_INCREMENT
Shouldn't this be ok? Why is it saying that he is finding tinyint?
这不应该好吗？为什么说他正在寻找tinyint？
采纳答案by Ian Dallas
Use @Basicfor basic integers. You can always trying declaring your ID as a Long though. I usually always use Long for my IDs. See Mapping Identifier Properities:
对基本整数使用@Basic。不过，您始终可以尝试将您的 ID 声明为 Long。我通常总是使用 Long 作为我的 ID。请参阅映射标识符属性：
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
回答by Pedro Ná?ez
I know this question is (really!) old, but:
我知道这个问题（真的！）很老，但是：
TINYINT represents 8-bit values. It's mapped to byte/Byte. It has a minimum value of -128 and a maximum value of 127 (inclusive) in both cases.
TINYINT 表示 8 位值。它映射到字节/字节。在这两种情况下，它的最小值为 -128，最大值为 127（含）。
SMALLINT represents 16-bit values. It's mapped to short/Short.
SMALLINT 表示 16 位值。它映射到短/短。
INTEGER represents 32-bit values. It's mapped to int/Integer.
INTEGER 表示 32 位值。它被映射到 int/Integer。
BIGINT represents 64-bit values. It's mapped to long/Long.
BIGINT 表示 64 位值。它被映射到 long/Long。
So, you can't map a tinyint using an Integer; you must use a Byte.
因此，您不能使用 Integer 映射 tinyint；你必须使用一个字节。
https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.htmlhttp://dev.mysql.com/doc/refman/5.7/en/integer-types.html
https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html http://dev.mysql.com/doc/refman/5.7/en/integer-types.html
回答by zakmck
I've just had this problem and it should be useful to share the details. What was happening in my case is:
我刚刚遇到了这个问题，分享详细信息应该很有用。在我的情况下发生的事情是：
- you define @Column (... definition = "integer" )
- Oracle translated 'integer' into decimal(0,-127) (or decimal(0), I haven't it clear)
- You use some other application where 'test.hibernate.hbm2ddl.auto' is 'validate'
- Hibernate reports that integer != decimal and validation fails
- 你定义 @Column (... definition = "integer" )
- Oracle 将'整数' 翻译成十进制(0,-127)（或十进制（0），我还不清楚）
- 您使用其他一些应用程序，其中“test.hibernate.hbm2ddl.auto”是“validate”
- Hibernate 报告 integer != decimal 和验证失败
my only solution has been to keep test.hibernate.hbm2ddl.auto empty (undefined = the DB is not touched). Not sure whether it's an Oracle or Hibernate bug (or both).
我唯一的解决方案是将 test.hibernate.hbm2ddl.auto 保持为空（未定义 = 未触及数据库）。不确定它是 Oracle 还是 Hibernate 错误（或两者兼有）。
回答by Sal
Add this to the annotations for the idfield:
将此添加到id字段的注释中：
@Column(columnDefinition = "TINYINT")
回答by bluesman
@Column(columnDefinition = "int")
回答by JamesB
Try adding the type annotation:
尝试添加类型注释：
@Type(type = "org.hibernate.type.IntegerType")
