使用scala / java调用存储过程oracle参数数组

时间:2016-06-23 01:11:23

标签: java oracle scala playframework

我有这段代码:

程序:

CREATE OR REPLACE PACKAGE BODY PKG_ARRAY AS   PROCEDURE PARAAJA_BULK(P_INPUTS IN PARAAJAARRAY) 
IS   BEGIN
        FOR I IN 1 .. P_INPUTS.COUNT LOOP

          INSERT INTO PARA_AJA
            (FIELD1, FIELD2)
          VALUES
            (P_INPUTS(I).FIELD1, P_INPUTS(I).FIELD2);
        END LOOP;   END;

    END;

和scala代码:

def spArray(name: List[person2],con:Connection) = Future[Boolean] { //supaya output jadi Future

    DriverManager.registerDriver(new OracleDriver())
    val conn = DriverManager.getConnection("jdbc:oracle:thin:@(DESCRIPTION=(ADDRESS=(PROTOCOL=TCP)(HOST=xxxx)(PORT=xxxx))(CONNECT_DATA=(SERVER=dedicated)(SERVICE_NAME=xxxx)))", "tesschema", "1234")
    var callableStatement: CallableStatement = conn.prepareCall("call PKG_ARRAY.PARAAJA_BULK(?)")

    val des = ArrayDescriptor.createDescriptor("PKG_ARRAY.PARAAJAARRAY", conn)
    val array_to_pass = new ARRAY(des, conn, name)

    callableStatement.setArray(1, array_to_pass)
    callableStatement.execute()

  }

但是,我有问题!!

无效的名称模式:PKG_ARRAY.PARAAJAARRAY

...

2 个答案:

答案 0 :(得分:1)

在java中我使用oracle过程如:

CallableStatement cstmt = Connection.prepareCall("{CALL CmdtyStndrd.CmdtyStndrdProc(?)}");
cstmt.registerOutParameter(1, array_to_pass);
cstmt.execute();

它有效。

答案 1 :(得分:0)

PARAAJAARRAY应该是模式级别定义的类型,而不是在包级别定义的类型。尝试使用CREATE OR REPLACE TYPE PARAAJAARRAY创建类型为TABLE OF(您的记录对象类型)。

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