python 3.6 mongoengine 无法捕获异常

Question

我有一个 django 1.9 项目，正在使用 mongoengine 连接到 mongodb 实例。

我有一个模型如下：

from django.db import models
from mongoengine import * 
class Employee(Document):
    email = StringField(required=True, unique=True)
    first_name = StringField(max_length=50)
    last_name = StringField(max_length=50)

测试class:

import web.models as models
class test:

    '''Test the accessibility of the business class'''
    def testBusiness():
        employee = models.Employee.objects.create(
            email="pedro.kong@company.com",
            first_name="Pedro",
            last_name="Kong"
        )
        try:
            employee.save()
        except pymongo.errors.DuplicateKeyError as e:
            return "Error"
        return "Business Tested"

问题是我无法捕获 DuplicateKeyError 执行：

pymongo.errors.DuplicateKeyError: E11000 duplicate key error index: rockynode.employee.$email_1  dup key: { : "pedro.kong@company.com" }

我尝试使用 global try except 但它仍然使代码崩溃并出现 DuplicateKeyError 异常：

try:
    employee.save()
except:
    return "Error"
return "Business Tested"

有关信息，我正在使用以下依赖项：

Django==1.9.7
mongoengine==0.10.6
pymongo==3.2.2

参考：get-the-duplicate-value-on-duplicatekeyerror

任何帮助都将是高度 appreciable.Thx。

Answer 1

我必须解决这个问题。

try except 没有工作，因为 DuplicateKeyError exception 不是从 save() 方法触发的，而是从 create() 方法触发的。

引用文档：create() :

Create new object. Returns the saved object instance.

mongoengine代码中def create()中的代码段：Reff:Monogoengine code清楚地表明create()函数调用了save()函数：

从 mongoengine 代码中提取：

def create(self, **kwargs):
        """Create new object. Returns the saved object instance.
        .. versionadded:: 0.4
        """
        return self._document(**kwargs).save()

因此我的 try 块必须包含 create() 方法，如下所示：

 try:
            employee = models.Employee.objects.create(
                email="pedro.kong@company.com",
                first_name="Pedro",
                last_name="Kong"
            )
            employee.save()
        except Exception as e:
            return "Error \n %s" % (e)
        return "Business Tested"

Answer 2

正如您在文档中看到的那样，create returns 一个保存的文档，所以如果您使用它

您不必再次保存

你必须在那里捕获保存异常

try:
    employee = models.Employee.objects.create(
        email="pedro.kong@company.com",
        first_name="Pedro",
        last_name="Kong"
    )
except Exception as e:
    return "Error \n %s" % (e)
return "Business Tested"

或者如果您不使用创建但保存在单独的语句中：

    employee = models.Employee(
        email="pedro.kong@company.com",
        first_name="Pedro",
        last_name="Kong"
    )
    try:
        employee.save()
    except Exception as e:
        return "Error \n %s" % (e)
    return "Business Tested"

python 3.6 mongoengine 无法捕获异常

python 3.6 mongoengine cannot catch exception

python

django

mongoengine

python-3.x

pymongo-3.x

从 mongoengine 代码中提取：