我刚开始学习MySQL和JDBC。
我使用phpmyadmin创建了一个名为testdb的表。表只有2列,分别为first和last。当我尝试从我的java类连接数据库时,我得到了MySQLSyntaxError。但是我无法理解。
这是我的班级:
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
public class Main {
public static void main(String[] args) throws ClassNotFoundException, SQLException {
String url = "jdbc:mysql://localhost:3306/testdb";
//Accessing driver from the JAR file.
Class.forName("com.mysql.jdbc.Driver");
//Creating a variable for the connection "con"
Connection con = DriverManager.getConnection(url,"root","password");
//Here is the query
PreparedStatement statement = con.prepareStatement("select * from name");
//Execute query
ResultSet result = statement.executeQuery();
while(result.next()) {
System.out.println(result.getString(1) + " " + result.getString(2));
}
}
}
以下是例外:
Exception in thread "main" com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Table 'testdb.name' doesn't exist
at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:57)
at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
at java.lang.reflect.Constructor.newInstance(Constructor.java:525)
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:1053)
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4096)
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4028)
at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:2490)
at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2651)
at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2734)
at com.mysql.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:2155)
at com.mysql.jdbc.PreparedStatement.executeQuery(PreparedStatement.java:2322)
at Main.main(Main.java:22)
感谢您的帮助
答案 0 :(得分:2)
MySQLSyntaxErrorException: Table 'testdb.name' doesn't exist
错误几乎是描述性的。没有名称和#34;名称"在testdb
架构中。
我使用phpmyadmin创建了一个名为 testdb 的表。
如果您已创建表testdb
,那么它应为select * from testdb
。不是吗?
答案 1 :(得分:2)
您已创建名为 testdb 的表格,因此您的查询应为
select * from testdb
不是select * from name
你应该检查你的堆栈痕迹。