Parameter index out of range (1 > number of parameters, which is 0)

遇到Parameter index out of range (1 > number of parameters, which is 0)怎么办？
2011年05月12日 星期四 23:45

遇到了Parameter index out of range (1 > number of parameters, which is 0) ，检查了很多遍错误。最后终于发现，原来PreparedStatement的sql中？是不能带有'单引号的。谢天谢地...一个很简单的失误，我折腾了4个钟头，忘大家引以为戒啊...

// 查询指定用户
public User_table selOperation(String id) {
   String sql = "select * from user_table where Id='?' ";
   User_table user = null;
   try {
    PreparedStatement pst = (PreparedStatement) AdoOpen.openConn()
      .prepareStatement(sql);
    pst.setString(1, id);
    ResultSet rs = pst.executeQuery();
    if (rs.next()) {
     user = new User_table();
     user.setId(rs.getString(1));
     user.setName(rs.getString(2));
     user.setPassword(rs.getString(3));
     user.setOnline(rs.getInt(4));
     user.setQita(rs.getString(5));

    }
    rs.close();
    pst.close();
    //AdoOpen.CloseConn();
   } catch (SQLException e) {
    e.printStackTrace();
   }

   return user;

}






java.sql.SQLException: Parameter index out of range (1 > number of parameters, which is 0).
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1055)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:956)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:926)
at com.mysql.jdbc.PreparedStatement.checkBounds(PreparedStatement.java:3326)
at com.mysql.jdbc.PreparedStatement.setInternal(PreparedStatement.java:3310)
at com.mysql.jdbc.PreparedStatement.setString(PreparedStatement.java:4146)
at ado.User_ADO.selOperation(User_ADO.java:60)
at unit.ProssSeverl.rundom(ProssSeverl.java:95)
at unit.ProssSeverl.regidit(ProssSeverl.java:71)
at unit.ProssSeverl.Dealwith(ProssSeverl.java:43)
at unit.ProssSeverl.service(ProssSeverl.java:36)
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:128)
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:286)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:845)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:583)
at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:447)
at java.lang.Thread.run(Unknown Source)


原来:


String sql = "select * from user_table where Id='?' ";


?不可以加单引号 改为


String sql = "select * from user_table where Id=? ";

评论

4 楼 q630077728 2016-03-11  

感谢楼主

3 楼 rj1989y 2012-10-21  

3q

2 楼 javarock 2012-06-13  

果然我也遇到了这个问题了 悲催！！！

1 楼 jiucool 2011-09-21  

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