2017-06-02 42 views
0

我在oracle(lro_dummy)中有一个自定义类型。我也有一个程序。我看起来像这样:cx_Oracle.OBJECT中的值缺失

procedure dummyLro 
(
    i_dummy   in lro_dummy, 
    o_dummy   out lro_dummy 
)is 
begin 
    o_dummy := lro_dummy('asdf'); 
end dummyLro; 

现在我想从python与cx_Oracle调用此过程。这看起来是这样的:

def test(db_con): 
    cur = db_con.cursor() 
    try: 
    procedure_params = {} 
    procedure_params["i_dummy"] = cur.var(cx_Oracle.OBJECT, typename="lro_dummy") 
    procedure_params["o_dummy"] = cur.var(cx_Oracle.OBJECT, typename="lro_dummy") 

    cur.callproc("test.dummyLro", [], procedure_params) 

o_dummy后具有价值,因为

str(procedures_params["o_dummy"]) 

回报

'<cx_Oracle.OBJECT with value <cx_Oracle.Object ???.LRO_DUMMY at 0x10492c9c0>>' 

但我不能访问我的属性。属性在

procedure_params["o_dummy"].type.attributes 

列出,但我无法找到

procedure_params["o_dummy"] 

我是怎么错的价值?

我使用Python 3.6cx_Oracle 5.3InstantClient 11.2Oracle-Server 11

+0

procedure_params [“o_dummy”] .type.attributes?中列出的属性是什么? –

+0

它返回包含一个项目的'cx_Oracle.ObjectAttribute DUMMY'数组。 name属性的值是'DUMMY' – Lee

+0

所以你应该可以通过procedure_params [“o_dummy”]来访问它。DUMMY? –

回答

0

我有同样的问题。根据“文档”(http://www.oracle.com/technetwork/articles/prez-stored-proc-084100.html),它应该按照您的操作方式工作,但它不适用于我。据我所知,传递给proc的params没有发生变化。而是返回新的对象。这对我有用:

i_dummy = cursor.var(cx_Oracle.OBJECT, typename='lro_dummy') 
o_dummy = cursor.var(cx_Oracle.OBJECT, typename='lro_dummy') 
[something, returned_o_dummy] = cur.callproc("test.dummyLro", [], [i_dummy, o_dummy]) 
print(returned_o_dummy.DUMMY)