如何映射JDBC

甲骨文對象我有這樣的Oracle對象：如何映射JDBC

CREATE OR REPLACE type employee_obj 
    AS 
     object (
     id NUMBER(10) , 
     ... 
    )

存儲過程

function get_employee_obj() return employee_obj is 
     l_employee  employee_obj; 
    begin 
     ... 
     return l_employee; 
    end;

，我需要從Java代碼調用它：

final String QUERY = "begin ? := GET_EMPLOYEE_OBJ(); end;"; 
    Connection connection = getConnection(); 

    CallableStatement stmt = connection.prepareCall(QUERY); 
    stmt.registerOutParameter(1, <WHAT TO PUT HERE>); 

    stmt.execute(); 
    ResultSet rs = (ResultSet) stmt.getObject(1);  
    ...

什麼SQL或Oracle類型我需要指定爲registerOutParameter參數從存儲函數讀取對象？我嘗試了幾次，但總是得到了PLS-00382：表達式是錯誤的類型錯誤。謝謝！

來源

2010-08-06 alex543

也許這將有助於與「se從雙重「get_employee_obj（）」：http://www.oracle.com/technology/sample_code/tech/java/sqlj_jdbc/files/advanced/ObjectOracleSample/ObjectOracleSample.java.html – 2010-08-06 09:52:37

http://download.oracle.com/文檔/ CD/B19306_01/java.102/b14355/oraoot.htm＃g1104293 – 2010-08-06 18:41:13

Oracle對象的正確類型是java.sql.Types.STRUCT或oracle.jdbc.OracleTypes.STRUCT，具體取決於您所需的支持級別。

但是擺脫錯誤PLS-00382: expression is of wrong type的訣竅是指定模式類型名稱。它應該拼寫爲，全部大寫，否則您將獲得invalid name pattern例外。

鑑於你的榜樣，這樣的事情可能是適當的：具有execute

stmt.registerOutParameter(1, OracleTypes.STRUCT, "EMPLOYEE_OBJ");

或

stmt.registerOutParameter(1, Types.STRUCT, "EMPLOYEE_OBJ");

後倒是查詢，您可以訪問結果爲：

STRUCT result = (oracle.sql.STRUCT)stmt.getObject(1); 
Object[] attr = result.getAttributes(); 
// attr[0] if the first field 
// attr[1] the second 
// and so on

來源

2014-09-07 20:56:25

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