Reputation: 807
when i was trying to read data from Oracle database using the following code i was getting exception
PreparedStatement pst=con.prepareStatement("SELECT * FROM TBLUSER");
pst.executeQuery();
But this table is actually exist in my database when i use this command directly in the command prompt its working fine.And also for one table among the tables in database this code is working fine ,but for other table names its not working properly.So someone please explain why this is happening.
java.sql.SQLException: ORA-00942: table or view does not exist
at oracle.jdbc.driver.DatabaseError.throwSqlException(DatabaseError.java:112)
at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:331)
at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:288)
at oracle.jdbc.driver.T4C8Oall.receive(T4C8Oall.java:743)
at oracle.jdbc.driver.T4CPreparedStatement.doOall8(T4CPreparedStatement.java:216)
at oracle.jdbc.driver.T4CPreparedStatement.executeForDescribe(T4CPreparedStatement.java:799)
at oracle.jdbc.driver.OracleStatement.executeMaybeDescribe(OracleStatement.java:1038)
at oracle.jdbc.driver.T4CPreparedStatement.executeMaybeDescribe(T4CPreparedStatement.java:839)
at oracle.jdbc.driver.OracleStatement.doExecuteWithTimeout(OracleStatement.java:1133)
at oracle.jdbc.driver.OraclePreparedStatement.executeInternal(OraclePreparedStatement.java:3285)
at oracle.jdbc.driver.OraclePreparedStatement.executeQuery(OraclePreparedStatement.java:3329)
at com.symp.ControllerServlet.service(ControllerServlet.java:302)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:727)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:303)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:208)
at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:241)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:208)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:220)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:122)
at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:501)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:170)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:98)
at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:950)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:116)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:408)
at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1040)
at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:607)
at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:315)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
at java.lang.Thread.run(Thread.java:745)
Upvotes: 0
Views: 22761
Reputation: 1367
I had this error when using persist() in Hibernate and the solution was to execute:
grant all on <schema>.<table> to <user>;
grant all on <schema>.<sequence> to <user>;
Upvotes: 0
Reputation: 807
the problem is i am using multiple connections in the program to login to application i have used Oracle connection which is given as class level variable.And the client will be accessed his tables present in MySQL database using MySQL connection which is service method level variable.By mistakenly i have referred oracle connection variable and tried to acess mysql table
Upvotes: 0
Reputation: 437
Check if you are using the correcte informations :
String url = "jdbc:mysql://localhost:3306/YOUR_SCHEMA";
Driver driver = com.mysql.jdbc.Driver();
String userName = "user";
String password = "pass";
Upvotes: 0
Reputation: 317
Check each of the following:
1) the spelling of the table or view name.
2) that a view is not specified where a table is required.
3) that an existing table or view name exists.
Contact the database administrator if the table needs to be created or if user or application privileges are required to access the table.
Upvotes: 1