Question

我是JDBC的初学者，我正试图在表格中插入一条记录，就像我创建raja一样：

所以这是我的计划：

import java.sql.*;

public class raj{
    public static void main(String[] args){
        try{
            Connection con=DriverManager.getConnection(
                    "jdbc:mysql://localhost:9999/raj1", "root", "1234"
            );
            Statement stmt=con.createStatement();

            String sql="insert into raja"+
                    "values(1, 'Rajendra Arora', 22)";
            stmt.executeUpdate(sql);

            sql="insert into raja"+
                    "values(2, 'Reena arora', 40)";
            stmt.executeUpdate(sql);

            sql="insert into raja"+
                    "values(3, 'Manohar lal arora', 45)";
            stmt.executeUpdate(sql);

            sql="insert into raja"+
                    "values(4, 'Suman and sudha arora', 20)";
            stmt.executeUpdate(sql);

            System.out.println("table records inserted succesfully....");

        }catch(Exception e){e.printStackTrace();}
    }
}

但在执行期间，显示的错误如下：

"C:\Program Files\Java\jdk1.8.0\bin\java" -Didea.launcher.port=7539 "-Didea.launcher.bin.path=C:\Program Files\JetBrains\IntelliJ IDEA 13.1.1\bin" -Dfile.encoding=UTF-8 -classpath "C:\Program Files\Java\jdk1.8.0\jre\lib\charsets.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\deploy.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\javaws.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\jce.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\jfr.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\jfxswt.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\jsse.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\management-agent.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\plugin.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\resources.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\rt.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\access-bridge.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\cldrdata.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\dnsns.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\jaccess.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\jfxrt.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\localedata.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\nashorn.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\sunec.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\sunjce_provider.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\sunmscapi.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\sunpkcs11.jar;C:\Program Files\Java\jdk1.8.0\jre\lib\ext\zipfs.jar;C:\Users\Raj Arora\IdeaProjects\untitled\out\production\untitled;C:\Users\Raj Arora\.IntelliJIdea13\config\jdbc-drivers\mysql-connector-java-5.1.29-bin.jar;C:\Program Files\JetBrains\IntelliJ IDEA 13.1.1\lib\idea_rt.jar" com.intellij.rt.execution.application.AppMain raj
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1, 'Rajendra Arora', 22)' at line 1
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
    at java.lang.reflect.Constructor.newInstance(Constructor.java:408)
    at com.mysql.jdbc.Util.handleNewInstance(Util.java:411)
    at com.mysql.jdbc.Util.getInstance(Util.java:386)
    at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1054)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4237)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4169)
    at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:2617)
    at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2778)
    at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2828)
    at com.mysql.jdbc.StatementImpl.executeUpdate(StatementImpl.java:1842)
    at com.mysql.jdbc.StatementImpl.executeUpdate(StatementImpl.java:1764)
    at raj.main(raj.java:13)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:483)
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:134)

Process finished with exit code 0

Answer 1

永远不要忘记空格：

String sql="insert into raja(col1,col2,col3) "+
                    "values(1, 'Rajendra Arora', 22)";

Answer 2

查看您的SQL请求！：

sql="insert into raja"+" values(3, 'Manohar lal arora', 45)";

添加空格！

Answer 3

我认为你需要添加一个＆＃34;白色空间＆＃34;介于＆＃34; raja＆＃34;和＆＃34;价值观＆＃34;。

此外，调试这些问题的最佳方法是在控制台中打印最终的sql并在任何数据库编辑器中使用它。

Answer 4

在你的问题中你提到你的sql如

String sql="insert into raja"+
                "values(1, 'Rajendra Arora', 22)";

但在

 raja"+ "values

是 rajavalues。

必须 raja值

所以把它作为

String sql="insert into raja "+
                "values(1, 'Rajendra Arora', 22)";

给我MySql语法错误？

4 个答案: