blocks|key|353976|text|我似乎还记得，为了让Hibernate在PostgreSQL上使用“序列”列，我不得不使用@GeneratedValue(strategy+=+GenerationType.IDENTITY)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|353977|entityMap^0|19|1F|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

I seem to recall having to use <code>@GeneratedValue(strategy = GenerationType.IDENTITY)</code> to get Hibernate to use 'serial' columns on PostgreSQL.

blocks|key|569742|text|我认为你已经有了足够的答案，但我得到了完全相同的错误，我的问题是另一个。我浪费了一点时间来解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|569743|在我的例子中，问题出在Postgres中序列的所有者。因此，如果上面的任何解决方案都不能解决您的问题，请检查序列的所有者是否是应该拥有权限的用户/角色。|569744|下面是一个示例：|569745|CREATE+SEQUENCE+seq_abcd
++++START+WITH+1
++++INCREMENT+BY+1
++++NO+MINVALUE
++++NO+MAXVALUE
++++CACHE+1;

ALTER+TABLE+public.seq_abcd+OWNER+TO+USER_APP;|code-block|syntax|javascript|569746|我希望它对任何人都有用。|569747|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I think you already have enough answer, but I got exactly the same error and my problem was another one. And I wasted a little bit of time trying to solve it.

In my case the problem was the owner of sequence in Postgres. So, if any solution above did not solved your problem, check if the owner of sequence is the user/role which should have permission.

Follows a sample:

<pre><code>CREATE SEQUENCE seq_abcd
 START WITH 1
 INCREMENT BY 1
 NO MINVALUE
 NO MAXVALUE
 CACHE 1;

ALTER TABLE public.seq_abcd OWNER TO USER_APP;
</code></pre>

I hope it can be useful for anyone.

blocks|key|354100|text|我之前遇到过同样的错误，请在您的数据库CREATE+SEQUENCE+hibernate_sequence+START+WITH+1+INCREMENT+BY+1+NOCYCLE;中键入以下查询|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|354101|这是我的工作，祝你好运~|354102|entityMap^0|J|1Z|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

I got same error before,
type this query in your database <code>CREATE SEQUENCE hibernate_sequence START WITH 1 INCREMENT BY 1 NOCYCLE;</code>

that's work for me, good luck ~

blocks|key|354160|text|您需要使用策略GenerationType.IDENTITY而不是GenerationType.AUTO来设置@GeneratedId列|type|unstyled|depth|inlineStyleRanges|entityRanges|data|354161|@Id
@GeneratedValue(strategy+=+GenerationType.IDENTITY)
@Column(name+=+"JUD_ID")
private+Long+_judId;|code-block|syntax|javascript|354162|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You need to set your @GeneratedId column with strategy GenerationType.IDENTITY instead of GenerationType.AUTO

<pre><code>@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "JUD_ID")
private Long _judId;
</code></pre>

blocks|key|354204|text|将GeneratedValue和GenericGenerator与native策略配合使用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|354205|@Id
@GeneratedValue(strategy+=+GenerationType.SEQUENCE,+generator+=+"id_native")
@GenericGenerator(name+=+"id_native",+strategy+=+"native")
@Column(name+=+"id",+updatable+=+false,+nullable+=+false)
private+Long+id;|code-block|syntax|javascript|354206|我不得不创建一个名为hibernate_sequence的序列，因为Hibernate默认会查找这样的序列：|354207|create+sequence+hibernate_sequence+start+with+1+increment+by+50;
grant+usage,+select+on+all+sequences+in+schema+public+to+my_user_name;|354208|entityMap^0|1|E|G|G|X|6|0|0|A|I|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

Using the <code>GeneratedValue</code> and <code>GenericGenerator</code> with the <code>native</code> strategy:

<pre><code>@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "id_native")
@GenericGenerator(name = "id_native", strategy = "native")
@Column(name = "id", updatable = false, nullable = false)
private Long id;
</code></pre>

I had to create a sequence call <code>hibernate_sequence</code> as Hibernate looks up for such a sequence by default:

<pre><code>create sequence hibernate_sequence start with 1 increment by 50;
grant usage, select on all sequences in schema public to my_user_name;
</code></pre>

blocks|key|569922|text|如果使用Postgres，请手动创建名为'hibernate_+sequence‘的序列。看起来不错。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|569923|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

If using Postgres, create sequence manually with name 'hibernate_sequence'. It will work.

blocks|key|354270|text|请使用以下查询并更改您的表：CREATE+SEQUENCE+user_id_seq+START+1;+ALTER+TABLE+product.users+ALTER+COLUMN+user_id+SET+DEFAULT+nextval('user_id_seq');+ALTER+SEQUENCE+users.user_id_seq+OWNED+BY+users.user_id;|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|354271|并在实体类中使用this|354272|@GeneratedValue(strategy+=+GenerationType.SEQUENCE,generator="user_id_seq")|354273|entityMap^0|E|10|1F|2A|3Q|1K|0|0|0|23|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|K|$]]

Please use the following query and alter your table :
 <code>CREATE SEQUENCE user_id_seq START 1;</code>
<code>ALTER TABLE product.users ALTER COLUMN user_id SET DEFAULT nextval('user_id_seq');</code>
<code>ALTER SEQUENCE users.user_id_seq OWNED BY users.user_id;</code>

and use the this in your entity class

<code>@GeneratedValue(strategy = GenerationType.SEQUENCE,generator="user_id_seq")</code>

blocks|key|354332|text|对于任何使用FluentNHibernate+(我的版本是2.1.2)的人来说，它是一样的重复，但这是有效的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|354333|public+class+UserMap+:+ClassMap<User>
{
++++public+UserMap()
++++{
++++++++Table("users");
++++++++Id(x+=>+x.Id).Column("id").GeneratedBy.SequenceIdentity("users_id_seq");|code-block|syntax|javascript|354334|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For anyone using FluentNHibernate (my version is 2.1.2), it's just as repetitive but this works:

<pre><code>public class UserMap : ClassMap&lt;User&gt;
{
 public UserMap()
 {
 Table("users");
 Id(x =&gt; x.Id).Column("id").GeneratedBy.SequenceIdentity("users_id_seq");
</code></pre>

i have gwt application connect to postgres DB at the backend, and a java class 'Judgement' mapping the table 'judgements' in DB, when i tried to persistent a judgement into db, it threw the following errors:

<pre><code>Caused by: org.hibernate.exception.SQLGrammarException: could not get next sequence value
...
Caused by: org.postgresql.util.PSQLException: ERROR: relation "hibernate_sequence" does not exist
</code></pre>

my Judgement class looks like this

<pre><code>@Entity
@Table(name = "JUDGEMENTS")
public class Judgement implements Serializable, Cloneable {

 private static final long serialVersionUID = -7049957706738879274L;

 @Id
 @GeneratedValue(strategy = GenerationType.AUTO)
 @Column(name = "JUD_ID")
 private Long _judId;
...
</code></pre>

and my table judgements is: 

<pre><code> Column | Type | Modifiers 
-------------+-----------------------------+---------------------------------------------------------
 jud_id | bigint | not null default nextval('judgements_id_seq'::regclass)
 rating | character varying(255) | 
 last_update | timestamp without time zone | 
 user_id | character varying(255) | 
 id | integer | 
Indexes:
 "judgements_pkey" PRIMARY KEY, btree (jud_id)
Foreign-key constraints:
 "judgements_id_fkey" FOREIGN KEY (id) REFERENCES recommendations(id)
 "judgements_user_id_fkey" FOREIGN KEY (user_id) REFERENCES users(user_id)
</code></pre>

and i have a SEQUENCE name 'judgements_id_seq' in DB

can anyone tell me what's wrong??? thanks.

hibernate could not get next sequence value

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我让gwt应用程序在后端连接到postgres DB，并且在DB中有一个java类‘判断’映射表‘判断’，当我试图将判断持久化到数据库中时，它抛出了以下错误：Caused by: org.hibernate.exception.SQLGrammarException: could not get next sequen...

问hibernate无法获取下一个序列值
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