mysql – Play Framework CRUD搜索问题
内容导读
互联网集市收集整理的这篇技术教程文章主要介绍了mysql – Play Framework CRUD搜索问题,小编现在分享给大家,供广大互联网技能从业者学习和参考。文章包含4988字,纯文字阅读大概需要8分钟。
内容图文
![mysql – Play Framework CRUD搜索问题](/upload/InfoBanner/zyjiaocheng/907/76a3572ffbbd479297c441d474646c4c.jpg)
当我在Play的CRUD模块中尝试搜索某些实体时,我遇到了这个异常:
play.exceptions.JavaExecutionException: org.hibernate.exception.SQLGrammarException: could not execute query
at play.mvc.ActionInvoker.invoke(ActionInvoker.java:290)
at Invocation.HTTP Request(Play!)
Caused by: javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not execute query
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1235)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1168)
at org.hibernate.ejb.QueryImpl.getResultList(QueryImpl.java:250)
at play.db.jpa.JPAPlugin$JPAModelLoader.fetch(JPAPlugin.java:431)
at controllers.CRUD$ObjectType.findPage(CRUD.java:253)
at controllers.CRUD.list(CRUD.java:36)
at play.mvc.ActionInvoker.invokeControllerMethod(ActionInvoker.java:413)
at play.mvc.ActionInvoker.invokeControllerMethod(ActionInvoker.java:408)
at play.mvc.ActionInvoker.invoke(ActionInvoker.java:182)
... 1 more
Caused by: org.hibernate.exception.SQLGrammarException: could not execute query
at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:92)
at org.hibernate.exception.JDBCExceptionHelper.convert(JDBCExceptionHelper.java:66)
at org.hibernate.loader.Loader.doList(Loader.java:2452)
at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2192)
at org.hibernate.loader.Loader.list(Loader.java:2187)
at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:452)
at org.hibernate.hql.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:363)
at org.hibernate.engine.query.HQLQueryPlan.performList(HQLQueryPlan.java:196)
at org.hibernate.impl.SessionImpl.list(SessionImpl.java:1258)
at org.hibernate.impl.QueryImpl.list(QueryImpl.java:102)
at org.hibernate.ejb.QueryImpl.getResultList(QueryImpl.java:241)
... 7 more
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'CHARSET' in 'where clause'
at com.mysql.jdbc.Util.handleNewInstance(Util.java:409)
at com.mysql.jdbc.Util.getInstance(Util.java:384)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1054)
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3566)
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3498)
at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:1959)
at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2113)
at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2568)
at com.mysql.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:2113)
at com.mysql.jdbc.PreparedStatement.executeQuery(PreparedStatement.java:2275)
at org.hibernate.jdbc.AbstractBatcher.getResultSet(AbstractBatcher.java:208)
at org.hibernate.loader.Loader.getResultSet(Loader.java:1869)
at org.hibernate.loader.Loader.doQuery(Loader.java:718)
at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:270)
at org.hibernate.loader.Loader.doList(Loader.java:2449)
... 15 more
奇怪的是,搜索适用于某些实体,而不适用于其他实体.例如,对于以下实体,在搜索框中添加任何字符串都可以:
@Entity
public class Act extends Model {
@Transient
private static final int PAGE_SIZE = Integer.parseInt(Play.configuration.getProperty("pagination.size","10"));
@Required(message = "act.name.required")
public String name;
@Required(message = "act.description.required")
@Lob
@MaxSize(value=500, message="act.description.maxsize")
public String description;
public Blob image;
public boolean showInClosedMode;
@Temporal(TemporalType.TIMESTAMP)
public Date updated;
@Required(message = "act.theatre.required")
@ManyToOne
public Theatre theatre;
public boolean disabled;
@OneToMany(mappedBy="act")
public List<Offer> offers;
@ManyToMany(cascade=CascadeType.PERSIST)
public Set<Tag> tags;
@Transient
public String tagListSupport;
[... some methods ...]
}
但不是这个:
@Entity
public class User extends Model {
@Required(message = "user.name.required")
public String name;
@Email(message = "user.email.invalid")
public String email;
public String prefLang;
public Long prefCity;
public int credits;
public Date lastLogin;
@NoBinding("profile")
public boolean disabled;
@NoBinding("profile")
public boolean admin;
@NoBinding("profile")
public boolean anonymous;
@OneToMany(mappedBy = "owner")
public List<Ticket> tickets;
@ManyToMany
public List<Theatre> theatres;
public String googleId;
public String yahooId;
public String facebookId;
public String twitterId;
public String twitter_token;
public String twitter_secret;
public String username;
public String password;
@OneToMany(mappedBy = "user")
public List<Event> history;
[...Methods...]
}
知道为什么会这样吗?
解决方法:
我的猜测是你正在使用的数据库不喜欢你有一个名为“user”的表,这是你没有为你的表提供特定名称时得到的.我知道Postgres不允许“用户”表,因为“user”是一个关键字.我不确定MySQL.尝试在User类中的@Entity注释之后添加javax.persistence.Table注释:
@Entity
@Table(name = "my_user")
public class User extends Model {
...
}
在哪里你给它一个你想要的不是“用户”的名字.或者(但未经测试)您可以将“用户”名称包装在引号中:
@Entity
@Table(name = "\"user\"")
public class User extends Model {
...
}
内容总结
以上是互联网集市为您收集整理的mysql – Play Framework CRUD搜索问题全部内容,希望文章能够帮你解决mysql – Play Framework CRUD搜索问题所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错,欢迎将互联网集市网站推荐给程序员好友。
内容备注
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 gblab@vip.qq.com 举报,一经查实,本站将立刻删除。
内容手机端
扫描二维码推送至手机访问。