小编Dea*_*End的帖子

PSQLException:基于密码的身份验证

这是例外:当我在springs上运行我的testDao文件时,springs中是否有给数据库的路径?

    org.postgresql.util.PSQLException: The server requested password-based authentication, but no password was provided.
    at org.postgresql.core.v3.ConnectionFactoryImpl.doAuthentication(ConnectionFactoryImpl.java:473)
    at org.postgresql.core.v3.ConnectionFactoryImpl.openConnectionImpl(ConnectionFactoryImpl.java:203)
    at org.postgresql.core.ConnectionFactory.openConnection(ConnectionFactory.java:65)
    at org.postgresql.jdbc.PgConnection.<init>(PgConnection.java:159)
    at org.postgresql.Driver.makeConnection(Driver.java:416)
    at org.postgresql.Driver.connect(Driver.java:283)
    at java.sql.DriverManager.getConnection(Unknown Source)
    at java.sql.DriverManager.getConnection(Unknown Source)
    at com.test.dao.java.TestDao.getConnection(TestDao.java:37)
    at com.test.dao.java.TestDao.getTest(TestDao.java:61)
    at com.test.main.java.TestMain.main(TestMain.java:33)
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这里是我的>> testDao文件,与PostgreSQL建立连接

   @Component 
    public class TestDao {
    static PreparedStatement ps;
    ResultSet rs;
    Connection conn= null;

     /**
     * @return
     * @throws SQLException
     * @throws ClassNotFoundException
     */
    private Connection getConnection() throws SQLException, ClassNotFoundException,FileNotFoundException,NullPointerException{


         if(conn==null)
         {
        try {
            Class.forName("org.postgresql.Driver");
            conn = DriverManager.getConnection(
                       "jdbc:postgresql://localhost:5432/testdb?user=postgres & password=postgres");
                    conn.close();
        } …
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java postgresql spring

7
推荐指数
1
解决办法
1万
查看次数

为什么我们将setInt与select查询一起使用,而不是在数据库中已存在值时使用getInt

为什么我们使用setInt和select查询而不是在数据库中已存在值时使用getInt?

    try {
        conn = getConnection();


        ps = conn.prepareStatement("SELECT * FROM circle where id =?");

        ps.setInt(1, circleId);

        Circle circle = null;
        rs = ps.executeQuery();
        if (rs.next()) {
            //String s = rs.getString(circleId);

            circle = new Circle(circleId, rs.getString("name"));

        }
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java jdbc

-7
推荐指数
1
解决办法
136
查看次数

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