Question

try {
        Class.forName("org.apache.hive.jdbc.HiveDriver");
    } catch (ClassNotFoundException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
        System.exit(1);
    }
    String query = "SELECT spd_field_label_id FROM RAL WHERE SUBJECT_USER_ID = ?";

    PreparedStatement stmt = null;
    Connection con = null;
    boolean testCasePassed = false;

    try {

        con = DriverManager.getConnection("jdbc:hive2://localhost:10000/default", "", "");
        stmt = con.prepareStatement(query);

        stmt.setString(1, "USR-44");
        ResultSet resultSet = stmt.executeQuery(query);

        Assert.assertNotNull(resultSet);

        while (resultSet.next()) {
            testCasePassed = true;
            System.out.println("=======Test =========" + resultSet.getString("spd_field_label_id"));
        }
    } finally {

        if (stmt != null) {
            stmt.close();
        }

        if (con != null) {
            con.close();
        }

    }

    return testCasePassed;

RAL是一个简单的Hive表，其字符串类型为spd_field_label_id和SUBJECT_USER_ID。

使用Hive2简单的PreparedStatement抛出下面的错误堆栈跟踪。什么可能是错的任何指针？使用Statement而不是PreparedStatement并且不使用时，相同的查询工作正常？用于参数绑定。

org.apache.hive.service.cli.HiveSQLException: Error while compiling statement: FAILED: ParseException line 1:62 cannot recognize input near '?' '<EOF>' '<EOF>' in expression specification
at org.apache.hive.jdbc.Utils.verifySuccess(Utils.java:264)
at org.apache.hive.jdbc.Utils.verifySuccessWithInfo(Utils.java:250)
at org.apache.hive.jdbc.HiveStatement.runAsyncOnServer(HiveStatement.java:309)
at org.apache.hive.jdbc.HiveStatement.execute(HiveStatement.java:250)
at org.apache.hive.jdbc.HiveStatement.executeQuery(HiveStatement.java:434)

Answer 1

stmt.executeQuery(query);

您使用的是错误的方法。你已经准备好了这份声明。它已准备好执行。它应该是：

stmt.execute();

Hive 2 JDBC PreparedStatement抛出错误无法识别＆＃39;附近的输入？＆＃39; ＆＃39; <EOF>＆＃39; ＆＃39; <EOF>＆＃39;在表达规范中

1 个答案: