【发布时间】:2017-11-22 08:04:08
【问题描述】:
我有这个方法。
private final void updateAllTableFields(final Class clazz){
final String tableName = ((Table)clazz.getAnnotation(Table.class)).name();
final String sqlQuery = new StringBuilder("SET @ids = NULL; ")
.append("UPDATE ")
.append(tableName)
.append(' ')
.append("set activeRecord=:activeRecord ")
.append("where activeRecord=true and updateable=true ")
.append("and (SELECT @ids \\:= CONCAT_WS(',', id, @ids)); ")
.append("select @ids;")
.toString();
final Query query = session.createSQLQuery(sqlQuery)
.setParameter("activeRecord",Boolean.FALSE);
final Object idsList=query.uniqueResult();
System.out.println("idsList = " + idsList);
}
我想进行更新并返回受影响的 Id 这工作完美使用 rawSQL 以字符串方式返回 id 但我无法使用 Hibernate 任何提示使其工作!!!
在此先致谢并致以最诚挚的问候。
更新
我需要进行更新并返回受影响的 ID!我不想做一个简单的更新。
你可以check it out the original question here pal: https://stackoverflow.com/questions/44604763/java-hibernate-tips-about-update-all-table-fields-performance
更新 错误是
at org.hibernate.exception.internal.SQLExceptionTypeDelegate.convert(SQLExceptionTypeDelegate.java:80)
at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:126)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:112)
at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:89)
at org.hibernate.loader.Loader.getResultSet(Loader.java:2065)
at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1862)
at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1838)
at org.hibernate.loader.Loader.doQuery(Loader.java:909)
at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:354)
at org.hibernate.loader.Loader.doList(Loader.java:2553)
at org.hibernate.loader.Loader.doList(Loader.java:2539)
at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2369)
at org.hibernate.loader.Loader.list(Loader.java:2364)
at org.hibernate.loader.custom.CustomLoader.list(CustomLoader.java:353)
at org.hibernate.internal.SessionImpl.listCustomQuery(SessionImpl.java:1873)
at org.hibernate.internal.AbstractSessionImpl.list(AbstractSessionImpl.java:311)
at org.hibernate.internal.SQLQueryImpl.list(SQLQueryImpl.java:141)
at org.hibernate.internal.AbstractQueryImpl.uniqueResult(AbstractQueryImpl.java:966)
at company.nuevemil.code.finalizarEntornoDePrueba(Test.java:56)
at company.nuevemil.code.main(Test.java:27)
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'UPDATE student set activeRecord=false,uid=1 where activeRecord=true at line 1
【问题讨论】:
-
介意发布您在尝试时遇到的错误吗?
-
已经完成了,非常感谢
-
请在您提供的代码中提供变量
activeRecord的值。如果只是true,那么您提供的代码确实没有生成您提供的错误消息。
标签: java mysql sql hibernate select