我能够执行以下代码来从表中选择所有数据: -
package com.jdbc;
import java.sql.*;
//in kdb+3.x and above
//init table with
//\p 5001
//Employees:([]id:0 1 2;firstName:`Charlie`Arthur`Simon;lastName:`Skelton`Whitney`Garland;age:10 20 30;timestamp:.z.p+til 3)
public class Selection{
static final String JDBC_DRIVER="jdbc";
static final String DB_URL="jdbc:q:localhost:5001";
static final String USER="";
static final String PASS="";
public static void main(String[] args){
Connection conn=null;
Statement stmt=null;
try{
Class.forName(JDBC_DRIVER);
System.out.println("Connecting to database...");
conn=DriverManager.getConnection(DB_URL,USER,PASS);
System.out.println("Creating statement...");
stmt=conn.createStatement();
ResultSet rs=stmt.executeQuery("SELECT id, firstName, lastName, age,timestamp FROM Employees");
while(rs.next()){
long id=rs.getLong("id");
long age=rs.getLong("age");
String first=rs.getString("firstName");
String last=rs.getString("lastName");
Timestamp timestamp=rs.getTimestamp("timestamp");
System.out.print("ID: "+id);
System.out.print(", Age: "+age);
System.out.print(", FirstName: "+first);
System.out.println(", LastName: "+last);
System.out.println(", Timestamp: "+timestamp);
}
rs.close();
stmt.close();
conn.close();
}catch(SQLException se){
se.printStackTrace();
}catch(Exception e){
e.printStackTrace();
}finally{
try{
if(stmt!=null)
stmt.close();
}catch(SQLException se2){
}
try{
if(conn!=null)
conn.close();
}catch(SQLException se){
se.printStackTrace();
}
}
}
}
但每当我尝试运行其他查询时,代码就会给出错误,例如,对于以下代码: -
package com.jdbc;
import java.sql.*;
//in kdb+3.x and above
//init table with
//\p 5001
//Employees:([]id:0 1 2;firstName:`Charlie`Arthur`Simon;lastName:`Skelton`Whitney`Garland;age:10 20 30;timestamp:.z.p+til 3)
public class Insertion {
static final String JDBC_DRIVER="jdbc";
static final String DB_URL="jdbc:q:localhost:5001";
static final String USER="";
static final String PASS="";
public static void main(String[] args){
Connection conn=null;
Statement stmt=null;
try{
Class.forName(JDBC_DRIVER);
System.out.println("Connecting to database...");
conn=DriverManager.getConnection(DB_URL,USER,PASS);
System.out.println("Creating statement...");
stmt=conn.createStatement();
stmt.executeUpdate("INSERT INTO Employees VALUES(9, 10, 'X', 'Y', " + new java.sql.Timestamp(0) + ")");
stmt.close();
conn.close();
}catch(SQLException se){
se.printStackTrace();
}catch(Exception e){
e.printStackTrace();
}finally{
try{
if(stmt!=null)
stmt.close();
}catch(SQLException se2){
}
try{
if(conn!=null)
conn.close();
}catch(SQLException se){
se.printStackTrace();
}
}
}
}
对于此代码,我收到以下错误: -
连接数据库...创建语句... java.sql.SQLException: 在jdbc.q(jdbc.java:22)jdbc $ co.ex(jdbc.java:26)at jdbc $ st.executeUpdate(jdbc.java:89)at com.jdbc.Insertion.main(Insertion.java:27)
当我尝试使用聚合函数进行选择查询时,我用来获得类似的错误。总而言之,我只能做简单的数据选择,而不能做任何事情。
任何线索?
是否有任何研究;关于如何使用JDBC在KDB数据库上进行复杂查询的链接?
我也试过使用预备语句(代码如下),但仍然没有运气: -
package com.jdbc;
import java.sql.*;
//in kdb+3.x and above
//init table with
//\p 5001
//Employees:([]id:0 1 2;firstName:`Charlie`Arthur`Simon;lastName:`Skelton`Whitney`Garland;age:10 20 30;timestamp:.z.p+til 3)
public class Insertion {
static final String JDBC_DRIVER="jdbc";
static final String DB_URL="jdbc:q:localhost:5001";
static final String USER="";
static final String PASS="";
public static void main(String[] args){
Connection conn=null;
PreparedStatement stmt=null;
try{
Class.forName(JDBC_DRIVER);
System.out.println("Connecting to database...");
conn=DriverManager.getConnection(DB_URL,USER,PASS);
System.out.println("Creating statement...");
stmt=conn.prepareStatement("INSERT INTO Employees(id,firstName,lastName,age,timestamp) VALUES(?, ?, ?, ?, ?)");
stmt.setInt(1, 10);
stmt.setString(2, "X");
stmt.setString(3, "Y");
stmt.setInt(4, 5);
stmt.setTimestamp(5, new Timestamp(0));
stmt.executeUpdate();
stmt.close();
conn.close();
}catch(SQLException se){
se.printStackTrace();
}catch(Exception e){
e.printStackTrace();
}finally{
try{
if(stmt!=null)
stmt.close();
}catch(SQLException se2){
}
try{
if(conn!=null)
conn.close();
}catch(SQLException se){
se.printStackTrace();
}
}
}
}
答案 0 :(得分:0)
尝试使用本机API而不是JDBC。
答案 1 :(得分:0)
我一直在努力解决同样的INSERT问题。通过纯粹的机会,发现如果你改变了“价值观”。对于小写的价值观' ,然后 static INSERT语句开始工作。尽管如此,PreparedStatement执行仍然没有运气。 (就我而言,不幸的是,在非选项中使用本机API。)