我有一个模块A
,它有一个2个1和这个2个1
partner_user = fields.Many2one('res.partner', string='Hesaby Customer')
subscription_manager_id = fields.Many2one('n_hesaby_subscription_manager.subscription_manager', string='Subscription Manager ID')
在模块B
中,我如何知道模块A是否已经记录了伙伴用户等同于创建的伙伴用户,然后减少记录的A
id
created_by = fields.Many2one('res.users', string='Created By', default=lambda self: self.env.uid, readonly=True)
n_company = fields.Many2one('n_hesaby_subscription_manager.subscription_manager', string='Company')
a我不知道如何开始解决这个问题。你能帮我吗?我已经找了两天的方法了
这是我的电脑
@api.depends('created_by')
def _compute_user_compnay(self):
for n_record in self:
user_compnay = self.env['res.users'].search([('id', '=', lambda self: self.env.user.id])], limit=1)
print('user id i think',user_compnay)
user_compnay = self.env['n_hesaby_subscription_manager.subscription_manager'].search([('subscription_manager_lines', 'in', user_compnay)], limit=1)
n_record.name = result
created_by = fields.Many2one('res.users', string='Created By', default=lambda self: self.env.uid, readonly=True)
n_company = fields.Many2one('n_hesaby_subscription_manager.subscription_manager', string='Company',compute='_compute_user_compnay')
模块A
class n_subscription_manager(models.Model):
_name = 'n_hesaby_subscription_manager.subscription_manager'
_description = 'Hesaby subscription manager'
_inherit = ['mail.thread', 'mail.activity.mixin']
_columns = {
'subscription_manager_lines': fields.One2many('n_hesaby_subscription_manager.subscription_manager.lines','subscription_manager_id', string='Subscription Manager Lines'),
#Other Columns
}
subscription_manager_lines = fields.One2many('n_hesaby_subscription_manager.subscription_manager.lines','subscription_manager_id',track_visibility="always", string='Subscription Manager Lines')
class n_subscription_manager_lines(models.Model):
_name = 'n_hesaby_subscription_manager.subscription_manager.lines'
_description = 'Hesaby subscription manager lines'
_columns = {
'subscription_manager_id': fields.Many2one('n_hesaby_subscription_manager.subscription_manager', string='Subscription Manager ID'),
}
print('OUTPUT ok what')
@api.depends('hesaby_user')
def _compute_user_of_contact(self):
print('OUTPUT test')
for n_use in self:
result = None
user = self.env['res.users'].search([('partner_id', '=', n_use.hesaby_user.id)], limit=1)
print('OUTPUT',user)
result = user.id
n_use.odoo_user = result
# n_subscription.subscription_type = n_subscription.partner_id.subscription_type_contact = result
hesaby_user = fields.Many2one('res.partner', string='Hesaby Customer')
odoo_user = fields.Many2one('res.users', string='Odoo User', compute='_compute_user_of_contact')
user_rank = fields.Selection(string='User Rank', selection=[
('primary_user', 'Primary User'),
('user', 'User'),
])
subscription_manager_id = fields.Many2one('n_hesaby_subscription_manager.subscription_manager', string='Subscription Manager ID')
模块B
class n_hesaby_snap(models.Model):
_name = 'n_hesaby_snap.n_hesaby_snap'
_inherit = ['mail.thread', 'mail.activity.mixin']
_description = 'Hesaby Snap'
@api.depends('created_by')
def _compute_user_compnay(self):
for n_record in self:
print('OK!')
obj = self.env['n_hesaby_subscription_manager.subscription_manager']
obj.search([('subscription_manager_lines.hesaby_user ', '=', n_record.created_by.partner_id)])
n_record.n_company = obj
created_by = fields.Many2one('res.users', string='Created By', default=lambda self: self.env.uid, readonly=True)
n_company = fields.Many2one('n_hesaby_subscription_manager.subscription_manager', string='Company',compute='_compute_user_compnay')
A
记录是在用户注册为公司创建新联系人并将用户联系人添加到该联系人时创建的
B
如果用户是用户,则他将在网站上以A
行访问它
因此,我希望在B
中创建一条记录时,将n_company
计算到记录的A
中,该记录中包含用户
将
id
与lambda表达式(对函数的引用)进行比较时,搜索方法将返回一个空记录集使用
self.env.user
而不是使用搜索来获取相同的记录partner_user
是一个res.partner
记录,而created_by
是一个res.users
记录,您无法比较它们,可能您的意思是如果created_by
用户的相关伙伴等于partner_user
编辑:
您应该将搜索结果分配给
n_company
,而不是obj
您可以在一行中找到许多
A
记录,其中包含该用户。应该有另一个条件来过滤记录以仅获取一条记录下面的示例使用默认顺序获取第一条记录
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