Reputation: 4552
I am using JPA and c3p0 and attempting to query a table and getting back a stack trace claiming that the table doesn't exist. I can open a connection to the db in, for example, DbVisualizer, and see the table there. In fact, the debug statements from my app show it is able to make a connection and test its viability. But then it is not finding the table.
15:45:53.940 [http-8080-1] DEBUG o.h.e.j.i.LogicalConnectionImpl - Obtaining JDBC connection
15:45:53.940 [http-8080-1] DEBUG c.m.v.c.i.C3P0PooledConnectionPool - Testing PooledConnection [com.mchange.v2.c3p0.impl.NewPooledConnection@4d687dcd] on CHECKOUT.
15:45:53.949 [http-8080-1] DEBUG c.m.v.c.i.C3P0PooledConnectionPool - Test of PooledConnection [com.mchange.v2.c3p0.impl.NewPooledConnection@4d687dcd] on CHECKOUT has SUCCEEDED.
15:45:53.950 [http-8080-1] DEBUG c.m.v.resourcepool.BasicResourcePool - trace com.mchange.v2.resourcepool.BasicResourcePool@7930ebb [managed: 3, unused: 2, excluded: 0] (e.g. com.mchange.v2.c3p0.impl.NewPooledConnection@3e30e173)
15:45:53.950 [http-8080-1] DEBUG o.h.e.j.i.LogicalConnectionImpl - Obtained JDBC connection
15:45:53.966 [http-8080-1] DEBUG org.hibernate.SQL - select alert0_.rrdb_key as rrdb1_0_, alert0_.date as date0_, alert0_.hostname as hostname0_, alert0_.message as message0_, alert0_.program as program0_ from reportsDb.alerts alert0_ where (alert0_.message not like '%Anomolous%') and (alert0_.message not like '%Requeue%')
Hibernate: select alert0_.rrdb_key as rrdb1_0_, alert0_.date as date0_, alert0_.hostname as hostname0_, alert0_.message as message0_, alert0_.program as program0_ from reportsDb.alerts alert0_ where (alert0_.message not like '%Anomolous%') and (alert0_.message not like '%Requeue%')
15:45:54.013 [http-8080-1] DEBUG c.m.v2.c3p0.impl.NewPooledConnection - com.mchange.v2.c3p0.impl.NewPooledConnection@4d687dcd handling a throwable.
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Table 'reportsDb.alerts' doesn't exist
at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method) ~[na:1.6.0_45]
at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:39) ~[na:1.6.0_45]
at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:27) ~[na:1.6.0_45]
at java.lang.reflect.Constructor.newInstance(Constructor.java:513) ~[na:1.6.0_45]
at com.mysql.jdbc.Util.handleNewInstance(Util.java:406) ~[mysql-connector-java-5.1.6.jar:na]
...
Here is persistence.xml (in /src/main/resources/META-INF):
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="1.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd">
<persistence-unit name="reportsDb" transaction-type="RESOURCE_LOCAL">
<description>Hibernate</description>
<class>com.pronto.mexp.common.entity.Alert</class>
</persistence-unit>
</persistence>
A subsection of applicationContext.xml:
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-2.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.1.xsd">
<bean id="jpaDialect" class="org.springframework.orm.jpa.vendor.HibernateJpaDialect"/>
<bean id="reportsDbEntityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="reportsDbDataSource" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="showSql" value="true"/>
<property name="generateDdl" value="false" />
<property name="databasePlatform" value="org.hibernate.dialect.MySQL5InnoDBDialect" />
</bean>
</property>
<property name="persistenceUnitName" value="reportsDb" />
<property name="jpaDialect" ref="jpaDialect"/>
</bean>
<bean id="reportsDbDataSource" class="com.mchange.v2.c3p0.ComboPooledDataSource" destroy-method="close">
<property name="driverClass" value="com.mysql.jdbc.Driver"/>
<!--<property name="jdbcUrl" value="jdbc:mysql://devdbrw01:3306/mexp"/>-->
<property name="jdbcUrl" value="jdbc:mysql://report101:3306/worker_events"/>
<property name="user" value="********"/>
<property name="password" value="********"/>
<property name="acquireRetryDelay" value="1000"/>
<property name="acquireRetryAttempts" value="4"/>
<property name="breakAfterAcquireFailure" value="false"/>
<property name="testConnectionOnCheckout" value="true"/>
<property name="maxConnectionAge" value="14400"/>
<property name="maxIdleTimeExcessConnections" value="1800"/>
</bean>
<!-- DAOs -->
<bean id="genericReportsDbDAO" class="com.pronto.mexp.common.dal.GenericReportsDbJPADAOImpl"/>
<bean id="alertJPADAO" class="com.pronto.mexp.dal.AlertJPADAOImpl" parent="genericReportsDbDAO"/>
</beans>
The thing I find suspicious is the part of the hibernate query where it tries to query select ... from reportsDb.alerts alert0_
- how do I confirm that "reportsDb" actually stands for my data source that I spec'd in applicationContext.xml?
ETA: The entity, Alert, looks like this:
@Entity
@Table(name = "alerts", catalog = "reportsDb")
public class Alert {
int rrdbKey;
String hostname = "";
String message = "";
String program = "";
Date date = new Date();
@javax.persistence.Column(name = "rrdb_key", nullable = false, insertable = false, updatable = false, length = 10, precision = 0)
@Id
public int getRrdbKey() {
return rrdbKey;
}
public void setRrdbKey(int rrdbKey) {
this.rrdbKey = rrdbKey;
}
@javax.persistence.Column(name = "hostname", nullable = false, insertable = false, updatable = false, length = 32, precision = 0)
@Basic
public String getHostname() {
return hostname;
}
public void setHostname(String hostname) {
this.hostname = hostname;
}
@javax.persistence.Column(name = "message", nullable = false, insertable = false, updatable = false, length = 128, precision = 0)
@Basic
public String getMessage() {
return message;
}
public void setMessage(String message) {
this.message = message;
}
@javax.persistence.Column(name = "program", nullable = true, insertable = false, updatable = false, length = 40, precision = 0)
@Basic
public String getProgram() {
return program;
}
public void setProgram(String program) {
this.program = program;
}
@javax.persistence.Column(name = "date", nullable = false, insertable = false, updatable = false, length = 19, precision = 0)
@Basic
public Date getDate() {
return date;
}
public void setDate(Date date) {
this.date = date;
}
}
Upvotes: 8
Views: 38105
Reputation: 167
I had the same issue, my mysql db was on windows but i moved it to linux which resulted in the mysql syntax of not recognizing the table. The cause was that mysql on windows is case insensitive and case sensitive on linux I was able to resolve this by adding :
lower_case_table_names=1
in my.cnf.
Also make sure to include
[mysqld]
at the beginning of my.cnf to avoid another error
"MySQL my.cnf file - Found option without preceding group"
Upvotes: 2
Reputation: 3606
From your entity definition, remove the catalog = 'reportsDb'
part, since it is being used to build the query like select from 'reportsDb.alerts'
.
Mysql doesn't use catalogs, AFAIK.
Upvotes: 5
Reputation: 1
Just add the property in hibernate.cfg.xml file then update the project then run the main method and the exception will go..thank you..
<session-factory><property name="hibernate.hbm2ddl.auto">create</property> </session-factory>
Upvotes: 0
Reputation: 11
In my case i mapped wrong dialect in cfg.xml, I am using MySQL57 db but used MySQL dialect. When I replaced correct dialect to MySQL57Dialect it worked.
Upvotes: 1
Reputation: 2513
In my case it was an issue of Hibernate converting my table to lower case.
My error was:
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Table 'Pluto.c_story' doesn't exist
Pluto is my db and C_Story is my table (NB: not c_story - lower case).
All I had to do was the following:
spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
Well, I hope this helps someone.
Upvotes: 4
Reputation: 1219
I also faced this error. In my case, I wrote the wrong database name in jdbc datasource class.
I wrote cms
jdbc:mysql://localhost:3306/cms
<bean id="jdbcDataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="com.mysql.jdbc.Driver" />
<property name="url" value="jdbc:mysql://localhost:3306/cms" />
<property name="username" value="root" />
<property name="password" value="" />
</bean>
Actually, the name of my database was hit_kcsl
jdbc:mysql://localhost:3306/hit_kcsl
<bean id="jdbcDataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="com.mysql.jdbc.Driver" />
<property name="url" value="jdbc:mysql://localhost:3306/hit_kcsl" />
<property name="username" value="root" />
<property name="password" value="" />
</bean>
Upvotes: 1
Reputation: 21
had a similar error on my code and I changed the persistence file
<properties>
<!-- Properties for Hibernate -->
<property name="hibernate.hbm2ddl.auto" value="create-drop" />
<property name="hibernate.show_sql" value="false" />
</properties>
to
<properties>
<!-- Properties for Hibernate -->
<property name="hibernate.hbm2ddl.auto" value="update" />
<property name="hibernate.show_sql" value="false" />
</properties>
replaced the "create-drop" with "update"
thanks
Upvotes: 1
Reputation: 10872
After exasperadetly eliminating every single occurence of table XYZ
in my code, I found the actual issue: XYZ
wasn't being referenced by JPA, but by an old, invalid mysql trigger. Maybe consider looking for the error outside of your code.
Upvotes: 3
Reputation: 9581
In my case the reason was because of left outer join on subquery with alias , that is working on sql editor but not on JDBCspring, so i removed the left outer join of subquery and replaced it with left outer with no subquery
Upvotes: 1
Reputation: 19898
We faced the same issue. There was one SQL query that didn't pass with an error like
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Table 'my_database_name.*' doesn't exist
But the query itself contained no my_database_name references or even * signs.
Comparing to other queries, we found the difference and added ORDER BY to query and error disappeared. Could be a hack around jdbc or c3p0 logics, but it worked for us.
Upvotes: 1
Reputation: 580
If the table really, really, does exist in mySQL, and your using Linux/Unix, and the error shows the table name in wrong/upper-case, then the issue is that table names in MySQL are case sensitive and hibernate is upper casing them. I'm using hibernate 4.3.
I just had this issue. Explanation here: lower_case_table_names=1
--edit-- In retrospect, it's probably better to find and change any @Table or hbm.xml references to match the database. I ran a wizard that generated an hbm.xml with uppercase names -- didn't realize that it was in my project until just now. I'll leave this here to make people aware of the case sensitivity.
--end of edit--
Here is how I fixed it:
Add this to /etc/mysql/my.conf:
set lower_case_table_names=1 #(default value '0').
Upvotes: 7