当我尝试将带有 hibernate 的字段插入表中时出错
[英]Error when I try to insert fields with hibernate into a table
问题描述
javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not execute statement org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:154) org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:181) org.hibernate.query.internal.AbstractProducedQuery.executeUpdate(AbstractProducedQuery.java:1617) Util.SQLControl.insertOrderLine(SQLControl.java:66) Servlet.ShoppingCardServlet.processRequest(ShoppingCardServlet.java:53) Servlet.ShoppingCardServlet.doPost(ShoppingCardServlet.java:81) javax.servlet.http.HttpServlet.service(HttpServlet.java:660) javax.servlet.Z80791B3AE70
javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not execute statement org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:154) org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:181) org.hibernate.query.internal.AbstractProducedQuery.executeUpdate(AbstractProducedQuery.java:1617) Util.SQLControl.insertOrderLine(SQLControl.java:66) Servlet.ShoppingCardServlet.processRequest(ShoppingCardServlet.java:53) Servlet.ShoppingCardServlet.doPost(ShoppingCardServlet. java:81) javax.servlet.http.HttpServlet.service(HttpServlet.java:6607servlet.B)3AE. 02CB88C246876D9FAA8F8Z.HttpServlet.service(HttpServlet.java:741) org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
Root Cause:根本原因：
org.hibernate.exception.SQLGrammarException: could not execute statement org.hibernate.exception.internal.SQLExceptionTypeDelegate.convert(SQLExceptionTypeDelegate.java:63) org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:42) org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:113) org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:99) org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.executeUpdate(ResultSetReturnImpl.java:178) org.hibernate.engine.query.spi.N
Root Cause:根本原因：
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'pepsi' in 'field list' java.base/jdk.internal.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method) java.base/jdk.internal.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62) java.base/jdk.internal.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45) java.base/java.lang.reflect.Constructor.newInstance(Constructor.java:490) com.mysql.jdbc.Util.handleNewInstance(Util.java:406) com
Note The full stack trace of the root cause is available in the server logs.注意服务器日志中提供了根本原因的完整堆栈跟踪。
public void insertOrderLine(String orderName, String tableId, String pID, String pName , String category, int quantity, int price, String date){
Session session = HibernateUtil.getSessionFactory().openSession();
session.beginTransaction();
Query query = session.createSQLQuery("INSERT INTO "+ orderName+" VALUES ("+tableId +","+ pID+" ,"+pName +
","+category+","+quantity+ ","+ price+" ,"+ date+")");
int result = query.executeUpdate();
}
1 个解决方案
解决方案1
0 2019-11-11 15:21:10
You dont seem to be setting the columns to insert your values into, when you execute an insert query you need to specify exactly which columns to insert into, like so您似乎没有设置要插入值的列，当您执行插入查询时，您需要准确指定要插入的列，如下所示
INSERT INTO table([column1], [column1]) VALUES(value_for_col1, value_for_col2).
Im guessing thatws why hibernate is complaining about unknown columns.我猜这就是 hibernate 抱怨未知列的原因。 just guessing from your input parameters i imagine your query will need to look like this:只是从您的输入参数猜测我想您的查询将需要如下所示：
Query query = session.createSQLQuery("INSERT INTO "+ orderName+" ([orderName], [tableId], [pID], [pName], [category], [quantity], [price], [date])VALUES ("+tableId +","+ pID+" ,"+pName +
","+category+","+quantity+ ","+ price+" ,"+ date+")");
Also for complex queries i suggest storing them within a sql file, this just makes editing easier and makes the code a bit neater.同样对于复杂的查询，我建议将它们存储在 sql 文件中，这只会使编辑更容易并使代码更整洁。 then just load the sql file as a string.然后只需将 sql 文件作为字符串加载。
An even easier method to save in hibernate would just be to create an entity object, (create a class then annotate with @Entity ).保存在 hibernate 中的更简单的方法是创建一个实体 object，（创建一个 class 然后用@Entity注释）。 Hibernate will then save this object for you avoiding the need for native update queries!然后 Hibernate 将保存此 object 为您避免本机更新查询的需要！
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