我正在使用 Tomcat 6 和 Postgresql 8.4。我的代码如下所示:
try {
// Prepared statement inserting something...
} catch (final PSQLException e) {
LOG.log(Level.SEVERE, "Saving failed.", e);
} catch (final SQLException e) {
LOG.log(Level.SEVERE, "Saving failed (SQL).", e);
}
这个插入 DB 可能会导致我想用第一个catch
. 但是,我在日志中实际发现的是:
SEVERE: Saving failed (SQL).
org.postgresql.util.PSQLException: ERROR: duplicate key value violates unique constraint "UZIVATELIA_U_LOGIN"
at org.postgresql.core.v3.QueryExecutorImpl.receiveErrorResponse(QueryExecutorImpl.java:2102)
at org.postgresql.core.v3.QueryExecutorImpl.processResults(QueryExecutorImpl.java:1835)
at org.postgresql.core.v3.QueryExecutorImpl.execute(QueryExecutorImpl.java:257)
at org.postgresql.jdbc2.AbstractJdbc2Statement.execute(AbstractJdbc2Statement.java:500)
at org.postgresql.jdbc2.AbstractJdbc2Statement.executeWithFlags(AbstractJdbc2Statement.java:388)
at org.postgresql.jdbc2.AbstractJdbc2Statement.executeQuery(AbstractJdbc2Statement.java:273)
at org.apache.commons.dbcp.DelegatingPreparedStatement.executeQuery(DelegatingPreparedStatement.java:96)
at org.apache.commons.dbcp.DelegatingPreparedStatement.executeQuery(DelegatingPreparedStatement.java:96)
at cz.ecoremedia.realpad.web.backend.Users.saveNewUser(Users.java:119)
at cz.ecoremedia.realpad.web.backend.Users.saveUser(Users.java:237)
at org.apache.jsp.User_jsp._jspService(User_jsp.java:159)
at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:388)
at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:313)
at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:260)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:293)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:859)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:602)
at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
at java.lang.Thread.run(Thread.java:662)
如果我检查e
我捕获的类型,例如这样:
} catch (final SQLException e) {
LOG.log(Level.SEVERE, e.getClass().getName());
LOG.log(Level.SEVERE, "Saving failed (SQL).", e);
它仍然告诉我它是org.postgresql.util.PSQLException
。
当我在我的 Eclipse 中本地尝试这个时,它工作得很好——在第一个块PSQLException
中被正确捕获。catch
如果我正确理解了这个问题,那么I catch 是一个与实际抛出org.postgresql.util.PSQLException
的不同的类。org.postgresql.util.PSQLException
这怎么可能?我究竟做错了什么?
谢谢你的想法。