java.sql.SQLException: After end of result set in mysql

Question

I am trying to Download image (.png) file from MYSQL. some time it works fine.unable find exact problem. it works properly on Jboss server. throws an error while trying to run in my local machine on Apche.

Please help me to FIX the error. Here is my java code.

    try {
        conection = SQLUtil.createConnection(Constant.DataSourceName);
        st = conection.prepareStatement("SELECT image FROM TABLE_NAME WHERE Userid="+ getUserId());
        result = st.executeQuery();
        result.next();

        if(!result.next()){
        input = result.getAsciiStream(1);
        }
        FileOutputStream output = new FileOutputStream(getSignatureImageDestinationPath());
        int ch = input.read();
        while (ch != -1) {
            output.write((char) ch);
            ch = input.read();
        }
        output.close();
        input.close();
        result.close();
        SQLUtil.closeConnection(conection);
    } catch (Exception e) {
        System.out.println(e+":Error");
    } finally {
        if (st != null) {
            SQLUtil.closeConnection(conection);
        }
    }

Here is Stack trace output:

java.sql.SQLException: After end of result set
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1074)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:988)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:974)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:919)
at com.mysql.jdbc.ResultSetImpl.checkRowPos(ResultSetImpl.java:854)
at com.mysql.jdbc.ResultSetImpl.getAsciiStream(ResultSetImpl.java:1275)
at org.apache.tomcat.dbcp.dbcp.DelegatingResultSet.getAsciiStream(DelegatingResultSet.java:253)
at org.apache.tomcat.dbcp.dbcp.DelegatingResultSet.getAsciiStream(DelegatingResultSet.java:253)
at com.ninenexus.model.Signature.getSignatureImage(Signature.java:173)
at com.ninenexus.servlets.SaveMySignature.mySignatureDisplay(SaveMySignature.java:69)
at com.ninenexus.servlets.SaveMySignature.doPost(SaveMySignature.java:35)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:641)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:722)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:306)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:240)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:161)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:164)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:100)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:118)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:380)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:243)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:188)
at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:288)
at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
at java.lang.Thread.run(Unknown Source)

What's the error, specifically, when it doesn't work? Be sure to close all resources in the finally block. Also, why are you using characters to represent binary data? — bdkosher
– bdkosher, Commented May 7, 2013 at 11:50
Possibly not related to your problem, but your use of result.next is unusual. If your query returns more than one row then "input" will be left at whatever its initial value is. — DaveH
– DaveH, Commented May 7, 2013 at 11:54

Aashray · Accepted Answer · 2013-05-07 12:05:12Z

15

You are calling result.next() twice. I'm assuming that your query returns only 1 row since you are trying to match by Userid. When the second result.next() is being called, there is no row to be returned in the ResultSet. This is why an SQLException is being thrown. Remove the 1st result.next() like so:

result = st.executeQuery();
if(!result.next()){
    input = result.getAsciiStream(1);
    }

answered May 7, 2013 at 12:05

Aashray

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4 Comments

DaveH Over a year ago

That still won't work will it? It's saying if NOT result.next().

Aashray Over a year ago

Doesn't matter, the call itself will trigger the exception.

DaveH Over a year ago

OK, I can see that the exception will be throw by the second call. But lets say he removes the first call to result.next(). The remaining call will return true, meaning that the input variable will not be populated won't it?

Aashray Over a year ago

Logically, I don't know what he's trying to do, but it won't enter the if block if the result set is not empty.

Kick Buttowski · Accepted Answer · 2016-06-13 06:38:13Z

2

The second result.next is moving you past the end of the result set.

I think you want

result = st.executeQuery();
if(result.next()){
   input = result.getAsciiStream(1);
}

edited Jun 13, 2016 at 6:38

Kick Buttowski

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answered May 7, 2013 at 12:12

DaveH

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