Question

有没有办法在Django中获得后端中立的字典光标？这将是一个dict而不是一个元组的游标。我被迫使用Oracle进行我正在进行的学校项目。

在Python的MySQLDb模块中，它被称为DictCursor。

有了WoLpH鼓舞人心的建议，我知道我非常接近......

def dict_cursor(cursor):
    for row in cursor:
        yield dict(zip(cursor.description, row))

迭代并打印用于导致的每一行光标：

(482072, 602592, 1)
(656680, 820855, 2)
(574968, 718712, 4)
(557532, 696918, 3))

但是用dict_cursor我得到了：

{('NET_SPENT', <type 'cx_Oracle.NUMBER'>, 127, 22, 0, 0, 1): 482072, ('LOT', <type 'cx_Oracle.NUMBER'>, 12, 22, 11, 0, 0): 1, ('NET_COLLECTED', <type 'cx_Oracle.NUMBER'>, 127, 22, 0, 0, 1): 602592}
{('NET_SPENT', <type 'cx_Oracle.NUMBER'>, 127, 22, 0, 0, 1): 656680, ('LOT', <type 'cx_Oracle.NUMBER'>, 12, 22, 11, 0, 0): 2, ('NET_COLLECTED', <type 'cx_Oracle.NUMBER'>, 127, 22, 0, 0, 1): 820855}
{('NET_SPENT', <type 'cx_Oracle.NUMBER'>, 127, 22, 0, 0, 1): 574968, ('LOT', <type 'cx_Oracle.NUMBER'>, 12, 22, 11, 0, 0): 4, ('NET_COLLECTED', <type 'cx_Oracle.NUMBER'>, 127, 22, 0, 0, 1): 718712}
{('NET_SPENT', <type 'cx_Oracle.NUMBER'>, 127, 22, 0, 0, 1): 557532, ('LOT', <type 'cx_Oracle.NUMBER'>, 12, 22, 11, 0, 0): 3, ('NET_COLLECTED', <type 'cx_Oracle.NUMBER'>, 127, 22, 0, 0, 1): 696918}

我只想让它使用密钥，例如'NET SPENT'。

经过精炼之后，这似乎有效：

def dict_cursor(cursor):
    for row in cursor:
        out = {}
        for i,col in enumerate(cursor.description):
            out[col[0]] = row[i]
        yield out

-

{'NET_COLLECTED': 602592, 'NET_SPENT': 482072, 'LOT': 1}
{'NET_COLLECTED': 820855, 'NET_SPENT': 656680, 'LOT': 2}
{'NET_COLLECTED': 718712, 'NET_SPENT': 574968, 'LOT': 4}
{'NET_COLLECTED': 696918, 'NET_SPENT': 557532, 'LOT': 3}

Answer 1

你可以写几行：）

def dict_cursor(cursor):
    description = [x[0] for x in cursor.description]
    for row in cursor:
        yield dict(zip(description, row))

或者如果你真的想节省空间：

simplify_description = lambda cursor: [x[0] for x in cursor.description]
dict_cursor = lambda c, d: dict(zip(d, r) for r in c))

Django后端中立的DictCursor

1 个答案: