Python Django:“;类型为'的参数;int';是不合适的”;在oracle/base.py的_rowfactory中
我正在尝试将Django应用程序连接到Oracle DB,但查询结果出现异常:Python Django:“;类型为'的参数;int';是不合适的”;在oracle/base.py的_rowfactory中,python,django,oracle,apache,Python,Django,Oracle,Apache,我正在尝试将Django应用程序连接到Oracle DB,但查询结果出现异常: TypeError: argument of type 'int' is not iterable 异常位置:/opt/isep/venv/lib/python3.4/site-packages/django/db/backends/oracle/base.py,在第560行工厂中 def _rowfactory(row, cursor): # Cast numeric values as the appropri
TypeError: argument of type 'int' is not iterable
异常位置:/opt/isep/venv/lib/python3.4/site-packages/django/db/backends/oracle/base.py,在第560行工厂中
def _rowfactory(row, cursor):
# Cast numeric values as the appropriate Python type based upon the
# cursor description, and convert strings to unicode.
casted = []
for value, desc in zip(row, cursor.description):
if value is not None and desc[1] is Database.NUMBER:
precision = desc[4] or 0
scale = desc[5] or 0
if scale == -127:
if precision == 0:
# NUMBER column: decimal-precision floating point
# This will normally be an integer from a sequence,
# but it could be a decimal value.
if '.' in value:
value = decimal.Decimal(value)
else:
value = int(value)
else:
# FLOAT column: binary-precision floating point.
# This comes from FloatField columns.
value = float(value)
elif precision > 0:
# NUMBER(p,s) column: decimal-precision fixed point.
# This comes from IntField and DecimalField columns.
if scale == 0:
value = int(value)
else:
value = decimal.Decimal(value)
elif '.' in value:
# No type information. This normally comes from a
# mathematical expression in the SELECT list. Guess int
# or Decimal based on whether it has a decimal point.
value = decimal.Decimal(value)
else:
value = int(value)
elif desc[1] in (Database.STRING, Database.FIXED_CHAR,
Database.LONG_STRING):
value = to_unicode(value)
casted.append(value)
return tuple(casted)
值中的“if.”行出现错误,因为value=0
我使用的是:Python 3.4、Django 1.11、cx_Oracle 6.0
如何解决这个问题?好的,我已经移除了我的静脉,然后创建了另一个。这次我安装了Django 1.11.15和cx_Oracle 6.0。一切似乎都很顺利。异常消失。您能提供一个示例查询来进一步调查这个问题吗?只需简单的Django的Model.objects.all()